Text Book Back Questions and Answers - Chapter 4 - Transition and Inner Transition Elements - 12th Chemistry Guide Samacheer Kalvi Solutions

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Transition and Inner Transition Elements
Text BookEvalution
I. Choose the correct answer.
Questions 1.
Sc $(Z=21)$ is a transition element but Zinc $(z=30)$ is not because
(a) both $\mathrm{Sc}^{3+}$ and $\mathrm{Zn}^{2+}$ ions are colourless and form white compounds.
(b) in case of Sc, 3d orbital are partially filled but in $\mathrm{Zn}$ these are completely filled
(c) last electron as assumed to be added to $4 \mathrm{~s}$ level in case of zinc
(d) both $\mathrm{Sc}$ and $\mathrm{Zn}$ do not exhibit variable oxidation states
Answer:
(c) in case of Sc, 3d orbital are partially filled but in $\mathrm{Zn}$ these are completely filled
Question 2.
Which of the following $\mathrm{d}$ block element has half filled penultimate $\mathrm{d}$ sub shell as well as half filled valence sub shell?
(a) $\mathrm{Cr}$
(b) $\mathrm{Pd}$
(c) Pt
(d) none of these
Answer:
(a) $\mathrm{Cr}$
Hint: $\mathrm{Cr} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^1$
Question 3.
Among the transition metals of $3 \mathrm{~d}$ series, the one that has highest negative $\left(\mathrm{M}^{2+} / \mathrm{M}\right)$ standard electrode potential is
(a) $\mathrm{Ti}$
(b) $\mathrm{Cu}$
(c) $\mathrm{Mn}$
(d) $\mathrm{Zn}$
Answer:
(a) $\mathrm{Ti}$

Question 4.
Which one of the following ions has the same number of unpaired electrons as present in $\mathrm{V}^{3+}$ ?
(a) $\mathrm{Ti}^{3+}$
(b) $\mathrm{Fe}^{3+}$
(c) $\mathrm{Ni}^{2+}$
(d) $\mathrm{Cr}^{3+}$
Answer:
(c) $\mathrm{Ni}^{2+}$
Question 5.
The magnetic moment of $\mathrm{Mn}^{2+}$ ion is
(a) $5.92 \mathrm{BM}$
(b) $2.80 \mathrm{BM}$
(c) $8.95 \mathrm{BM}$
(d) $3.90 \mathrm{BM}$
Answer:
(a) $5.92 \mathrm{BM}$
Hint: $\mathrm{Mn}^{2+} \Rightarrow 3 \mathrm{~d}$ contains 5 unpaired electrons
$
\begin{aligned}
& \mathrm{n}=5 \\
& =\sqrt{n(n+2)} B M \\
& =\sqrt{5(5+2)}=\sqrt{35}=5.92 \mathrm{BM}
\end{aligned}
$
Question 6.
Which of the following compounds is colourless?
(a) $\mathrm{Fe}^{3+}$
(6) $\mathrm{Ti}^{4+}$
(c) $\mathrm{CO}^{2+}$
(d) $\mathrm{Ni}^2$
Answer:
(b) $\mathrm{Ti}^{4+}$
Hint: $\mathrm{Ti}^{4+}$ contains no unpaired electrons in d orbital, hence no d-d transition.
Question 7.
The catalytic behaviour of transition metals and their compounds is ascribed mainly due to
(a) their magnetic behaviour
(b) their unfilled d orbitals
(c) their ability to adopt variable oxidation states
(d) their chemical reactivity
Answer:
(c) their ability to adopt variable oxidation states

Question 8.
The correct order of increasing oxidizing power in the series
(a) $\mathrm{VO}_2^{+}<\mathrm{Cr}_2 \mathrm{O}_7^{2-}<\mathrm{MnO}_4^{-}$
(b) $\mathrm{Cr}_2 \mathrm{O}_7^{2-}<\mathrm{VO}_2^{+}<\mathrm{MnO}_4^{-}$
(c) $\mathrm{Cr}_2 \mathrm{O}_7^{2-}<\mathrm{MnO}_4^{-}<\mathrm{VO}_2^{+}$
(d) $\mathrm{MnO}_4^{-}<\mathrm{Cr}_2 \mathrm{O}_7^{2-}<\mathrm{VO}_2^{+}$
Answer:
(a) $\mathrm{VO}_2^{+}<\mathrm{Cr}_2 \mathrm{O}_7^{2-}<\mathrm{MnO}_4^{-}$
$+5 \quad+6 \quad+7$
Hint:

greater the oxidation state, higher is the oxidising power.
Question 9.
The alloy of copper that contain Zinc is
(a) Monel metal
(b) Bronze
(c) bell metal
(d) brass
Answer:
(d) brass
Question 10.
Which of the following does not give oxygen on heating?
(a) $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$
(b) $\left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7$
(c) $\mathrm{KClO}_3$
(d) $\mathrm{Zn}\left(\mathrm{ClO}_3\right)_2$
Answer:
(b) $\left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7$
Hint:
$
\begin{aligned}
& 2 \mathrm{~K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \stackrel{\Delta}{\longrightarrow} 2 \mathrm{~K}_2 \mathrm{CrO}_4+\mathrm{Cr}_2 \mathrm{O}_3+\left(\frac{3}{2}\right) \mathrm{O}_2 \\
& \left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7 \stackrel{\Delta}{\longrightarrow} \mathrm{Cr}_2 \mathrm{O}_3+4 \mathrm{H}_2 \mathrm{O}+\mathrm{N}_2 \\
& \mathrm{KClO}_3 \stackrel{\Delta}{\longrightarrow} \mathrm{KCl}++\left(\frac{3}{2}\right) \mathrm{O}_2 \\
& \mathrm{Zn}\left(\mathrm{ClO}_3\right) \stackrel{\Delta}{\longrightarrow} \mathrm{ZnCl}_2+3 \mathrm{O}_2
\end{aligned}
$

Question 11.
In acid medium, potassium permanganate oxidizes oxalic acid to
(a) Oxalate
(b) Carbon dioxide
(c) acetate
(d) acetic acid
Answer:
(b) Carbon dioxide
Hint: $5(\mathrm{COO})_2^{2-}+2 \mathrm{MnO}_4^{-}+16 \mathrm{H}^{+} \stackrel{\triangle}{\rightarrow} 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}^{2+} \uparrow+8 \mathrm{H}^2 \mathrm{O}$
Question 12 .
Which of the following statements is not true?
(a) on passing $\mathrm{H}_2 \mathrm{~S}$, through acidified $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ solution, a milky colour is observed.
(b) $\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is preferred over $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ in volumetric analysis
(c) $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ solution in acidic medium is orange in colour
(d) $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ solution becomes yellow on increasing the $\mathrm{pH}$ beyond 7
Answer:
(b) $\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is preferred over $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ in volumetric analysis
Question 13.
Permanganate ion changes to in acidic medium
(a) $\mathrm{MnO}_4^{2-}$
(b) $\mathrm{Mn}^{2+}$
(c) $\mathrm{Mn}^{3+}$
(d) $\mathrm{MnO}_2$
Answer:
(b) $\mathrm{Mn}^{2+}$
Hint: $\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}$
Question 14.
A white crystalline salt $(\mathrm{A})$ react with dilute $\mathrm{HCl}$ to liberate a suffocating gas (B) and also forms a yellow precipitate. The gas (B) turns potassium dichromate acidified with dil $\mathrm{H}_2 \mathrm{SO}_4$ to a green coloured solution(C). $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are respectively
(a) $\mathrm{Na}_2 \mathrm{SO}_3, \mathrm{SO}_2, \mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3$
(b) $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3, \mathrm{SO}_2, \mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3$
(c) $\mathrm{Na}_2 \mathrm{~S}, \mathrm{SO}_2, \mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3$
(d) $\mathrm{Na}_2 \mathrm{SO}_4, \mathrm{SO}_2, \mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3$

Answer:
(a) $\mathrm{Na}_2 \mathrm{SO}_3, \mathrm{SO}_2, \mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3$
Hint :

Question 15.
$\mathrm{MnO}_4^{-}$react with $\mathrm{Br}$ in alkaline $\mathrm{P}^{\mathrm{H}}$ to give
(a) $\mathrm{BrO}_3^{-}, \mathrm{MnO}_2$
(b) $\mathrm{Br}_3, \mathrm{MnO}_4^{2-}$
(c) $\mathrm{Br}_3, \mathrm{MnO}_3$
(d) $\mathrm{BrO}^{-}, \mathrm{MnO}_4^{2-}$
Answer:
(a) $\mathrm{BrO}_3^{-}, \mathrm{MnO}_2$
Hint: $2 \mathrm{MnO}_4^{-}+\mathrm{Br}-+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{OH}-+2 \mathrm{MnO}_2+\mathrm{BrO}_3^{-}$
Question 16.
How many moles of $\mathrm{I}_2$ are liberated when 1 mole of potassium dichromate react with potassium iodide?
(a) 1
(b) 2
(c) 3
(d) 4
Answer
(c) 3
Hint: $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+6 \mathrm{KI}+7 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow 4 \mathrm{~K}_2 \mathrm{SO}_4+\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+7 \mathrm{H}_2 \mathrm{O}+3 \mathrm{I}_2$
Question 17.
The number of moles of acidified $\mathrm{KMnO}_4$ required to oxidize 1 mole of ferrous oxalate $\left(\mathrm{FeC}_2 \mathrm{O}_4\right)$ is
(a) 5
(b) 3
(c) 0.6
(d) 1.5
Answer:
(c) 0.6
Hint: $\mathrm{MnO}_4^{-}+\mathrm{FeC}_2 \mathrm{O}_4 \rightarrow \mathrm{Mn}^{2+}+\mathrm{Fe}^{3+}+2 \mathrm{CO}_2$
$5 \mathrm{e}^{-}$acception $3 \mathrm{e}^{-}$release
5 moles of $\mathrm{FeC}_2 \mathrm{O}_4 \equiv 3$ moles of $\mathrm{KMnO}_4$
5 moles of $\mathrm{FeC}_2 \mathrm{O}_4 \equiv\left(\frac{3}{5}\right)$ moles of $\mathrm{KMnO}_4$
5 moles of $\mathrm{FeC}_2 \mathrm{O}_4 \equiv 0.6$ moles of $\mathrm{KMnO}_4$

Question 18.
When a brown compound of $\mathrm{Mn}(\mathrm{A})$ ids treated with $\mathrm{HCl}$, it gives a gas (B). The gas (B) taken in excess reacts with $\mathrm{NH}_3$ to give an explosive compound (C). The compound $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are
(a) $\mathrm{MnO}_2, \mathrm{Cl}_2, \mathrm{NCl}_3$
(b) $\mathrm{MnO}, \mathrm{Cl}_2, \mathrm{NH}_4 \mathrm{Cl}$
(c) $\mathrm{Mn}_3 \mathrm{O}_4, \mathrm{Cl}_2, \mathrm{NCl}_3$
(d) $\mathrm{MnO}_3, \mathrm{Cl}_2, \mathrm{NCl}_2$
Answer:
(a) $\mathrm{MnO}_2, \mathrm{Cl}_2, \mathrm{NCl}_3$

Question 19.
Which one of the following statements related to lanthanons is incorrect?
(a) Europium shows +2 oxidation state.
(b) The basicity decreases as the ionic radius decreases from $\mathrm{Pr}$ to $\mathrm{Lu}$.
(c) All the lanthanons are much more reactive than aluminium.
(d) $\mathrm{Ce}^{4+}$ solutions are widely used as oxidising agents in volumetric analysis.
Answer:
(c) All the lanthanons are much more reactive than aluminium.
Hint: As we move from La to $\mathrm{Lu}$, their metallic behaviour because almost similar to that of aluminium.
Question 20.
Which of the following lanthanoid ions is diamagnetic?
(a) $\mathrm{Eu}^{2+}$
(b) $\mathrm{Yb}^{2+}$
(c) $\mathrm{Ce}^{2+}$
(d) $\mathrm{Sm}^{2+}$
Answer:
(6) $\mathrm{Yb}^{2+}$
Hint: $\mathrm{Yb}^{2+}-4 \mathrm{f}^{14}$ - no unpaired electrons - diamagnetic

Question 21.
Which of the following oxidation states is most common among the lanthanoids?
(a) 4
(b) 2
(c) 5
(d) 3
Answer:
(d) 3
Question 22.
Assertion: $\mathrm{Ce}^{4+}$ is used as an oxidizing agent in volumetric analysis.
Reason: $\mathrm{Ce}^{4+}$ has the tendency of attaining +3 oxidation state.
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false. .
(d) Both assertion and reason are false.
Answer:
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
Question 23.
The most common oxidation state of actinoids is
(a) +2
(b) +3
(c) +4
(d) +6
Answer:
(c) +4
Question 24.
The actinoid elements which show the highest oxidation state of +7 are
(a) $\mathrm{Np}, \mathrm{Pu}, \mathrm{Am}$
(b) U, Fm, Th
(c) U, Th, Md
(d) Es, No, Lr
Answer:
(a) $\mathrm{Np}, \mathrm{Pu}, \mathrm{Am}$

Question 25.
Which one of the following is not correct?
(a) $\mathrm{La}(\mathrm{OH})_2$ is less basic than $\mathrm{Lu}(\mathrm{OH})_3$
(b) In lanthanoid series ionic radius of $\operatorname{Ln}^{3+}$ ions decreases
(c) La is actually an element of transition metal series rather than lanthanide series
(d) Atomic radii of $\mathrm{Zr}$ and $\mathrm{Hf}$ are same because of lanthanide contraction
Answer:
(a) $\mathrm{La}(\mathrm{OH})_2$ is less basic than $\mathrm{Lu}(\mathrm{OH})_3$
II. Answer the following Questions:
Question 1.
What are transition metals? Give four examples.
Answer:
1. Transition metal is an element whose atom has an incomplete $\mathrm{d}$ sub shell or which can give rise to cations with an incomplete $\mathrm{d}$ sub shell.
2. They occupy the central position of the periodic table, between s-block and p-block elements.
3. Their properties are transitional between highly reactive metals of s-block and elements of p-block which are mostly non metals.
4. Example - Iron, Copper, Tungsten, Titanium.
Question 2.
Explain the oxidation states of $3 \mathrm{~d}$ series elements.
Answer:
1. The first transition metal Scandium exhibits only +3 oxidation state, but all other transition elements exhibit variable oxidation states by losing electrons from (n-1)d orbital and ns orbital as the energy difference between them is very small.
2. At the beginning of the $3 \mathrm{~d}$ series, +3 oxidation state is stable but towards the end +2 oxidation state becomes stable.
3. The number of oxidation states increases with the number of electrons available, and it decreases as the number of paired electrons increases. For example, in $3 \mathrm{~d}$ series, first element Sc has only one oxidation state +3 ; the middle element Mn has six different oxidation states from +2 to +7 . The last element $\mathrm{Cu}$ shows +1 and +2 oxidation states only.
4. $\mathrm{Mn}^{2+}\left(3 \mathrm{~d}^5\right)$ is more stable than $\mathrm{Mn}^{4+}\left(3 \mathrm{~d}^3\right)$ is due to half filled stable configuration.
Question 3.
What are inner transition elements?
Answer:
1. The inner transition elements have two series of elements.
- Lanthanoids
- Actinoids

2. Lanthanoid series consists of 14 elements from Cerium $\left({ }_{58} \mathrm{Ce}\right)$ to Lutetium $\left({ }_{71} \mathrm{Lu}\right)$ following Lanthanum $\left.{ }_{57} \mathrm{La}\right)$. These elements are characterised by the preferential filling of $4 \mathrm{f}$ orbitals. .
3. Actinoids consists of 14 elements from Thorium $\left({ }_{90} \mathrm{Th}\right)$ to Lawrencium ( $\left.{ }_{103} \mathrm{Lr}\right)$ following Actinium ( 89 Ac). These elements are characterised by the preferential filling of $5 \mathrm{f}$ orbital.
Question 4.
Justify the position of lanthanides and actinides in the periodic table.
Answer:
1. In sixth period after lanthanum, the electrons are preferentially filled in inner $4 \mathrm{f}$ sub shell and these 14 elements following lanthanum show similar chemical properties. Therefore these elements are grouped together and placed at the bottom of the periodic table. This position can be justified as follows.
- Lanthanoids have general electronic configuration $[\mathrm{Xe}] 4 \mathrm{f}^{2-14} 5 \mathrm{~d}^{0-1} 6 \mathrm{~s}^2$
- The common oxidation state of lanthanoids is +3
- All these elements have similar physical and chemical properties.
2. Similarly the fourteen elements following actinium resemble in their physical and chemical properties.
3. If we place these elements after Lanthanum in the periodic table below $4 \mathrm{~d}$ series and actinides below $5 \mathrm{~d}$ series, the properties of the elements belongs to a group would be different and it would affect the proper structure of the periodic table.
4. Hence a separate position is provided to the inner transition elements at the bottom of the periodic table.
Question 5.
What are actinoides? Give three examples.
Answer:
1. The fourteen elements following actinium ,i.e., from thorium (Th) to lawrentium (Lr) are called actinoids.
2. All the actinoids are radioactive and most of them have short half lives.
3. The heavier members being extremely unstable and not of natural occurrence. They are produced synthetically by the artificial transformation of naturally occuring elements by nuclear reactions.
4. Example - Thorium, Uranium, Plutonium, Californium.
Question 6.
Why $\mathrm{Gd}^{3+}$ is colourless?
Answer:
$\mathrm{Gd}$ - Electronic Configuration: $[\mathrm{Xe}] 4 \mathrm{f}^7 5 \mathrm{~d}^1 6 \mathrm{~s}^2$
$\mathrm{Gd}^{3+}$ - Electronic Configuration: $[\mathrm{Xe}] 4 \mathrm{f}^7$
In $\mathrm{Gd}^{3+}$, no electrons are there in outer d-orbitals. d-d transition is not possible. So it is colourless.

Question 7.
Explain why compounds of $\mathrm{Cu}^{2+}$ are coloured but those of $\mathrm{Zn}^{2+}$ are colourless.

Answer:
$\mathrm{Cu}(\mathrm{Z}=29)$ Electronic configuration is $[\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^1$
$\mathrm{Cu}^{2+}$. Electronic configuration is $[\mathrm{Ar}] 3 \mathrm{~d}^9$.
In $\mathrm{Cu}^{2+}$, promotion of electrons take place in outer d-orbital by the absorption of light form visible region
involves d-d transition. Due to this $\mathrm{Cu}^{2+}$ compounds are coloured. Where in $\mathrm{Zn}^{2+}$ electronic configuration is $[\mathrm{Ar}] 3 \mathrm{~d}^{10}$. It has completely filled d-orbital. So there is no chance of $\mathrm{d}-\mathrm{d}$ transition. So $\mathrm{Zn}^{2+}$ compounds are colourless.
Question 8.
Describe the preparation of potassium dichromate.
Answer:
Preparation of potassium dichromate:
1. Potassium dichromate is prepared from chromite ore. The ore $\mathrm{FeO} \cdot \mathrm{Cr}_2 \mathrm{O}_3$ is concentrated by gravity separation process.
2. The concentrated ore is mixed with excess sodium carbonate and lime and roasted in a reverbratory furnace.
$
4 \mathrm{FeCr}_2 \mathrm{O}_4+8 \mathrm{Na}_2 \mathrm{CO}_3+7 \mathrm{O}_2 \underline{900-1000^0 C_{\longrightarrow}} 8 \mathrm{Na}_2 \mathrm{CrO}_4+2 \mathrm{Fe}_2 \mathrm{O}_3+8 \mathrm{CO}_2 \uparrow
$
3. The roasted mass is treated with water to separate soluble sodium chromate from insoluble iron oxide. The yellow solution of sodium chromate is treated with concentrated sulphuric acid which converts sodium chromate into sodium dichromate.

4. The above solution is concentrated to remove less soluble sodium sulphate. The resulting solution is filtered and concentrated. It is cooled to get the crystals of $\mathrm{Na}_2 \mathrm{SO}_2 \cdot 2 \mathrm{H}_2 \mathrm{O}$.
5. The saturated solution of sodium dichromate in water is mixed with $\mathrm{KCl}$ and then concentrated to get crystals of $\mathrm{NaCl}$. It is filtered while hot and the filtrate is cooled to obtain $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ crystals.

Question 9.
What is lanthanide contraction and what are the effects of lanthanide contraction?
Answer:
As we move across $4 \mathrm{f}$ series, the atomic and ionic radii of lanthanoids show gradual decrease with increase in atomic number. This decrease in ionic size is called lanthanoid contraction.
Effects (or) Consequences of lanthanoid contraction:
1. Basicity differences: As we move from $\mathrm{Ce}^{3+}$ to $\mathrm{Lu}^{3+}$, the basic character of $\mathrm{Ln}^{3+}$ ions decrease. Due to the decrease in the size of $\mathrm{Ln}^{3+}$ ions, the ionic character of $\mathrm{Ln}-\mathrm{OH}$ bond decreases (covalent character increases) which results in the decrease in the basicity.
2. Similarities among lanthanoids - In the complete f-series only $10 \mathrm{pm}$ decrease in atomic radii and $20 \mathrm{pm}$ decrease in ionic radii is observed. Because of this very small change in radii of lanthanoids, their chemical properties are quite similar.

The elements of second and third transition series resemble each other more closely than the elements of first and second transition series due to lanthanoid contraction. For example,
- $4 \mathrm{~d}$ series $-\mathrm{Zr}$ - Atomic radius $145 \mathrm{pm}$
- $5 \mathrm{~d}$ series - $\mathrm{Hf}-$ Atomic radius $144 \mathrm{pm}$
Question 10.
Complete the following

a. $\mathrm{MnO}_4{ }^{2-}+\mathrm{H}^{+} \longrightarrow$ ?
b. $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3 \frac{\text { acidified }}{\mathrm{KMnO}_4}$ ?
c. $\mathrm{MnO}_4^{-}+\mathrm{Fe}^{2+} \longrightarrow$ ?
d. $\mathrm{KMnO}_4 \frac{\Delta}{\text { Red hot }}$ ?
e. $\mathrm{Cr}_2 \mathrm{O}_7^{2-}+\mathrm{I}^{-}+\mathrm{H}^{+} \longrightarrow$ ?
f. $\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{KCl} \longrightarrow$ ?

Answer:

(a) 

(b) 

(c) 

(d) 

(e) 

(f) 

Question 11.
What are interstitial compounds?
Answer:
1. An interstitial compound or alloy is a compound that is formed when small atoms like hydrogen, boron, carbon or nitrogen are trapped in the interstitial holes in a metal lattice.
2. They are usually non-stoichiometric compounds.
3. Transition metals form a number of interstitial compounds such as $\mathrm{TiC}, \mathrm{ZrH}_{1.92}, \mathrm{Mn}_4 \mathrm{~N}$ etc.
4. The elements that occupy the metal lattice provide them new properties.
- They are hard and show electrical and thermal conductivity
- They have high melting points higher than those of pure metals
- Transition metal hydrides are used as powerful reducing agents
- Metallic carbides are chemically inert.
Question 12.
Calculate the number of unpaired electrons in $\mathrm{Ti}^{3+}, \mathrm{Mn}^{2+}$ and calculate the spin only magnetic moment.
Answer:
$\mathrm{Ti}^{3+}$
$\mathrm{Ti}(\mathrm{Z}=22)$. Electronic configuration $[\mathrm{Ar}] 3 \mathrm{~d}^2 4 \mathrm{~s}^2$
$\mathrm{Ti}^{3+}-$ Electronic configuration $[\mathrm{Ar}] 3 \mathrm{~d}^1$
So, the number of unpaired electrons in $\mathrm{Ti}^{3+}$ is equal to 1 .
Spin only magnetic moment of $\mathrm{Ti}^{3+}=\sqrt{1(1+2)}=\sqrt{3}=1.73 \mu_{\mathrm{B}}$
$\mathrm{Mn}^{2+}$
$\mathrm{Mn}(\mathrm{Z}=25)$. Electronic configuration $[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^2$
$\mathrm{Mn}^{2+}-$ Electronic configuration $[\mathrm{Ar}] 3 \mathrm{~d}^5$
$\mathrm{Mn}^{2+}$ has 5 unpaired electrons.
Spin only magnetic moment of $\mathrm{Mn}^{2+}=\sqrt{5(5+2)}=\sqrt{35}=5.91 \mu_{\mathrm{B}}$
Question 13.
Write the electronic configuration of $\mathrm{Ce}^{4+}$ and $\mathrm{CO}^{2+}$.
Answer:
$\mathrm{Ce}(\mathrm{Z}=58) \rightarrow \mathrm{Ce}^4+4 \mathrm{e}^{-}$
$\mathrm{Ce}^{4+}-\mathrm{Is}^2-2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 4 \mathrm{~s}^2 3 \mathrm{~d}^{10} 4 \mathrm{p}^6 5 \mathrm{~s}^2 4 \mathrm{~d}^{10} 5 \mathrm{p}^6$
$\mathrm{CO}^{2+}-\mathrm{Is}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 4 \mathrm{~s}^2 3 \mathrm{~d}^5$.

Question 14.
Explain briefly how +2 states becomes more and more stable in the first half of the first row transition elements with increasing atomic number.
First transition series.
Answer:

Question 15 .
Which is more stable? $\mathrm{Fe}^{3+}$ or $\mathrm{Fe}^{2+}-$ explain.
Answer:
$\mathrm{Fe}(\mathrm{Z}=26)$
$\mathrm{Fe} \rightarrow \mathrm{Fe}^{2+}+2 \mathrm{e}^{-}$
$\mathrm{Fe} \rightarrow \mathrm{Fe}^{3+}+3 \mathrm{e}^{-}$
$\mathrm{Fe}^{2+}$ [Number of electrons 24]
Electronic configuration $=[\mathrm{Ar}] 3 \mathrm{~d}^6$
$\mathrm{Fe}^{3+}$ [Number of electrons 23]
Electronic configuration $=[\mathrm{Ar}] 3 \mathrm{~d}^5$
Among $\mathrm{Fe}^{3+}$ and $\mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}$ is more stable due to half filled d-orbital. This can be explained by Aufbau principle. Half filled and completely filled d-orbitals are more stable than partially filled d-orbitals. So $\mathrm{Fe}^{3+}$ is more stable than $\mathrm{Fe}^{2+}$.
Question 16.

$\text { Explain the variation in } \mathrm{E}^0 \mathrm{M}^{2+} / \mathrm{M}^{3+3 \mathrm{~d}} \text { series. }$
Answer:
1. In transition series, as we move down from Ti to $\mathrm{Zn}$, the standard reduction potential $\mathrm{E}^0 \mathrm{M}^{2+} / \mathrm{M}^3$ value is approaching towards less negative value and copper has a positive reduction potential, i.e. elemental copper is more stable than $\mathrm{Cu}^{2+}$.
2. $\mathrm{E}^0 \mathrm{M}^{2+} / \mathrm{M}$ value for manganese and zinc are more negative than regular trend. It is due to extra stability arises due to the half filled $\mathrm{d}^5$ configuration in $\mathrm{Mn}^{2+}$ and completely filled $\mathrm{d}^{10}$ configuration in $\mathrm{Zn}^{2+}$.
3. The standard electrode potential for the $\mathrm{M}^{3+} / \mathrm{M}^{2+}$ half cell gives the relative stability between $\mathrm{M}^{3+}$ and $\mathrm{M}^{2+}$
4. The high reduction potential of $\mathrm{Mn}^{3+} / \mathrm{Mn}^{2+}$ indicates $\mathrm{Mn}^{2+}$ is more stable than $\mathrm{Mn}^{3+}$.
5. For $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$ the reduction potential is $0.77 \mathrm{~V}$, and this low value indicates that both $\mathrm{Fe}^{3+}$ and $\mathrm{Fe}^{2+}$ can exist under normal condition.
6. $\mathrm{Mn}^{3+}$ has a $3 \mathrm{~d}^2$ configuration while that of $\mathrm{Mn}^{2+}$ is $3 \mathrm{~d}^5$. The extra stability associated with a half filled d sub-shell makes the reduction of $\mathrm{Mn}^{3+}$ very feasible $\left[\mathrm{E}^{\circ}=+1.51 \mathrm{~V}\right]$
Question 17.
Compare lanthanides and actinides.
Answer:
Lanthanoids:
1. Differentiating electron enters in $4 \mathrm{f}$ orbital
2. Binding energy of $4 \mathrm{f}$ orbitals are higher
3. They show less tendency to form complexes
4. Most of the lanthanoids are colourless
5. They do not form oxo cations
6. Besides +3 oxidation states lanthanoids show +2 and +4 oxidation states in few cases.
Actinoids:
1. Differentiating electron enters in $5 \mathrm{f}$ orbital
2. Binding energy of $5 f$ orbitals are lower
3. They show greater tendency to form complexes
4. ost of the actinoids are coloured.
E.g: $\mathrm{U}^{3+}$ (red), $\mathrm{U}^{4+}$ (green), $\mathrm{UO}_2^{2+}$ (yellow)
5. They do form oxo cations such as $\mathrm{UO}_2^{2+} \mathrm{NpO}_2^{2+}$ etc.
6. Besides +3 oxidation states actinoids show higher oxidation states such as $+4,+5,+6$ and +7 .

Question 18.
Explain why $\mathrm{Cr}^{2+}$ is strongly reducing while $\mathrm{Mn}^{3+}$ is strongly oxidizing.
Answer:
$\mathrm{Cr}^{2+}$ is strong reducing while $\mathrm{Mn}^{3+}$ is strongly oxidising.
$
\begin{aligned}
& \mathrm{E}^0 \mathrm{Cr}^{3+} / \mathrm{Cr}^{2+} \text { is }-0.41 \mathrm{~V} \\
& \mathrm{Cr}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cr} \mathrm{E}=-0.91 \mathrm{~V} \text {. }
\end{aligned}
$
If the standard electrode potential $\mathrm{E}^{\circ}$ of a metal is large and negative, the metal is a powerful reducing agent because it loses electrons easily.
$
\begin{aligned}
& \mathrm{Mn} \rightarrow \mathrm{Mn}^{3+}+3 \mathrm{e}^{-} \\
& \mathrm{Mn}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} \\
& \mathrm{Mn}^{3+}[\mathrm{Ar}] 3 \mathrm{~d}^4 \\
& \mathrm{E}^{\circ}=+1.51 \mathrm{~V}
\end{aligned}
$
If the standard electrode potential $\mathrm{E}^{\circ}$ of a metal is large and positive, the metal is a powerful oxidising agent because it gains electrons easily.
Question 19.
Compare the ionization enthalpies of first series of the transition elements. Ionization enthalpies of first transition series:
Answer:
1. Ionization energy of transition element is intermediate between those of $s$ and $p$ block elements.
2. As we move from left to right in a transition metal series, the ionization enthalpy increases as expected. This is due to increase in nuclear charge corresponding to the filling of $\mathrm{d}$ electrons.
3. The increase in first ionisation enthalpy with increase in atomic number along a particular series is not regular. The added electron enters $(\mathrm{n}-1) \mathrm{d}$ orbital and the inner electrons act as a shield and decrease the effect of nuclear charge on valence 'ns' electrons. Therefore, it leads to variation in the ionization energy values.
Question 20.
Actinoid contraction is greater from element to element than the lanthanoid contraction, why? Answer:
1. Actinoid contraction is greater from element to element than lanthanoid contraction. The $5 f$ orbitals in Actinoids have a very poorer shielding effect than $4 \mathrm{f}$ orbitals in lanthanoids.
2. Thus, the effective nuclear charge experienced by electron in valence shells in case of actinoids is much more than that experienced by lanthanoids.
3. In actinoids, electrons are shielded by $5 \mathrm{~d}, 4 \mathrm{f}, 4 \mathrm{~d}$ and $3 \mathrm{~d}$ whereas in lanthanoids, electrons are shielded by $4 \mathrm{~d}, 4 \mathrm{f}$ only.
4. Hence, the size contraction in actinoids is greater as compared to that in lanthanoids.

Question 21.
Out of $\mathrm{LU}(\mathrm{OH})_3$ and $\mathrm{La}(\mathrm{OH})_3$ which is more basic and why?
Answer:
1. As we move from $\mathrm{Ce}^{3+}$ to $\mathrm{Lu}^{3+}$, the basic character of $\mathrm{Lu}^{3+}$ ions decreases.
2. Due to the decrease in the size of $\mathrm{Lu}^{3+}$ ions, the ionic character of $\mathrm{Lu}-\mathrm{OH}$ bond decreases, covalent character increases which results in the decrease in the basicity.
3. Hence, $\mathrm{La}(\mathrm{OH})^3$ is more basic than $\mathrm{Lu}(\mathrm{OH})^3$.
Question 22.
Why europium (II) is more stable than Cerium (II)?
Answer:
$\mathrm{Eu}(\mathrm{Z}=63)$ - Electronic configuration - $[\mathrm{Xe}] 4 \mathrm{f}^7 5 \mathrm{~d}^{\circ} 6 \mathrm{~s}^2$
$\mathrm{Eu}^{2+}-$ Electronic configuration Electronic $6 \mathrm{~s}^1$
$\mathrm{Ce}(\mathrm{Z}=58)$ - configuration - $[\mathrm{Xe}] 4 \mathrm{f}^{2+} 5 \mathrm{~d}^{\circ} 6 \mathrm{~s}^{2+}$
$\mathrm{Ce}^{2+}$ - Electronic confluration $-[\mathrm{Xe}] 4 \mathrm{f}^2 5 \mathrm{~d}^{\circ}$
According to Aufbau principle, half filled and completely filled d (or) f orbitals are more stable than partially filled f orbitals.
Hence $\mathrm{Eu}^{2+}[\mathrm{Xe}] 4 \mathrm{f}^7 5 \mathrm{~d}^{\circ}$ is more stable than $\mathrm{Ce}^{2+}[\mathrm{Xe}] 4 \mathrm{f}^2 5 \mathrm{~d}^{\circ}$
Question 23.
Why do zirconium and Hafnium exhibit similar properties?
Answer:
1. The element of second and third transition series resemble each other more closely than the elements of first and second transition series due to lanthanoid contraction.
2. e.g., $\mathrm{Zr}-4 \mathrm{~d}$ series -Atomic radius $145 \mathrm{pm} \mathrm{Hf}-5 \mathrm{~d}$ series - Atomic radius $144 \mathrm{pm}$
3. The radii are very similar even though the number of electrons increases.
4. $\mathrm{Zr}$ and $\mathrm{Hf}$ have very similar chemical behaviour, having closely similar radii and electronic configuration.
5. Radius dependent properties such as lattice energy, solvation energy are similar.
6. Thus lanthanides contraction leads to formation of pair of elements and those known as chemical twins, e.g., $\mathrm{Zr}-\mathrm{Hf}$
Question 24.
Which is stronger reducing agent $\mathrm{Cr}^{2+}$ or $\mathrm{Fe}^{2+}$ ?
Answer:
$\mathrm{Cr}^{2+}$ and $\mathrm{Fe}^{2+}$
$\mathrm{Cr}(\mathrm{Z}=24)-$ Electronic configuration $-[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^1$

$\mathrm{Cr}^{2+}$ Electronic configuration - $[\mathrm{Ar}] 3 \mathrm{~d}^4 4 \mathrm{~s}^0$
$\mathrm{Fe}(\mathrm{Z}=26)$ - Electronic configuration $-[\mathrm{Ar}] 3 \mathrm{~d}^6 4 \mathrm{~s}^2$
$\mathrm{Fe}^{2+}$ Electronic confimiration - $[\mathrm{Ar}] 3 \mathrm{~d}^6 4 \mathrm{~s}^0$
If standard electrode potential $\left(\mathrm{E}^{\circ}\right)$ of a metal is large and negative, the metal is a powerful reducing agent
$
\begin{aligned}
& \mathrm{Cr}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cr} \\
& \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \\
& \mathrm{E}^{\circ}=-0.91 \mathrm{~V} \\
& \mathrm{E}^{\circ}=-0.44 \mathrm{~V}
\end{aligned}
$
By comparing the above equation, $\mathrm{Cr}^{2+}$ is a powerful reducing agent.
Question 25.
The $\mathrm{E}^0{ }_{\mathrm{M}^{2+}}{ }_M$ value for copper is positive. Suggest a possible reason for this.
Answer:
1. Copper has a positive reduction potential. Elemental copper is more stable than $\mathrm{Cu}^{2+}$.
2. Copper having positive sign for electrode potential merely means that copper can undergo reduction at faster rate than reduction of hydrogen.
3. The electron giving reaction (oxidation) of copper is slower than that of hydrogen. It is determined from the result of S.H.E (Standard Hydrogen Electrode) potential value experiment.

Question 26.
Predict which of the following will be coloured in aqueous solution $\mathrm{Ti}^{2+}, \mathrm{V}^{3+}, \mathrm{Sc}^{4+}, \mathrm{Cu}^{+}, \mathrm{SC}^{3+}, \mathrm{Fe}^{3+}$, $\mathrm{Ni}^{2+}$ and $\mathrm{CO}^{3+}$
Answer:
Among $\mathrm{Ti}^{2+}, \mathrm{V}^{3+}, \mathrm{Sc}^{4+}, \mathrm{Cu}^{+}, \mathrm{Sc}^{3+}, \mathrm{Fe}^{3+}, \mathrm{Ni}^{2+}$ and $\mathrm{CO}^{3+}$ in aqueous solution state.
$\mathrm{Ti}(\mathrm{Z}=22)-\mathrm{Ti}^{2+}-$ Electronic configuration is $[\mathrm{Ar}] 3 \mathrm{~d}^2$
$\mathrm{V}(\mathrm{Z}=23)-\mathrm{V}^{3+}$ - Electronic configuration is $[\mathrm{Ar}] 3 \mathrm{~d}^2$
$\mathrm{Sc}(\mathrm{Z}=21)-\mathrm{SC}^{4+}$ - Electronic configuration is $[\mathrm{Ar}] 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^5$
$\mathrm{Cu}(\mathrm{Z}=29)-\mathrm{Cu}^{+}$- Electronic configuration is $[\mathrm{Ar}] 3 \mathrm{~d}^{10}$
$\mathrm{Sc}(\mathrm{Z}=21)-\mathrm{SC}^{3+}$ - Electronic configuration is $[\mathrm{Ar}] 3 \mathrm{~d}^{\circ} 4 \mathrm{~s}^{\circ}$
$\mathrm{Fe}(\mathrm{Z}=26)-\mathrm{Fe}^{3+}$ - Electronic configuration is $[\mathrm{Ar}] 3 \mathrm{~d}^5$
$\mathrm{Ni}(\mathrm{Z}=28)-\mathrm{Ni}^{2+}$ - Electronic configuration is $[\mathrm{Ar}] 3 \mathrm{~d}^5$
$\mathrm{CO}(\mathrm{Z}=27)-\mathrm{CO}^{3+}$ - Electronic configuration is $[\mathrm{Ar}] 3 \mathrm{~d}^6$
A transition metal ion is coloured if it has one or more unpaired electrons in (n-1)d orbital, i.e. 3d orbitals in the case of first transition series, when such species are exposed to visible radiation, $d-d$ transition take place and the species are coloured.

Question 27.
Describe the variable oxidation state of $3 \mathrm{~d}$ series elements.
Answer:
1. The first transition metal Scandium exhibits only +3 oxidation state, but all other transition elements exhibit variable oxidation states by loosing electrons from (n-1)d orbital and ns orbital as the energy difference between them is very small.
2. At the beginning of the $3 \mathrm{~d}$ series, +3 oxidation state is stable but towards the end +2 oxidation state becomes stable.
3. The number of oxidation states increases with the number of electrons available, and it decreases as the number of paired electrons increases. For example, in the $3 \mathrm{~d}$ series, first element Sc has only one oxidation
state +3 the middle element Mn has six different oxidation states from +2 to +7 . The last element Cu shows +1 and +2 oxidation states only.
4. $\mathrm{Mn}^{2+}\left(3 \mathrm{~d}^5\right)$ is more stable than $\mathrm{Mn}^{4+}\left(3 \mathrm{~d}^3\right)$ is due to half filled stable configuration.
Question 28.
Which metal in the $3 \mathrm{~d}$ series exhibits +1 oxidation state most frequently and why?
Answer:
1. The first transition metal copper exhibits only +1 oxidation state.
2. It is unique in $3 \mathrm{~d}$ series having a stable +1 oxidation state.
$\mathrm{Cu}(\mathrm{Z}=29)$ Electronic configuration is $[\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^2$
3. So copper element only can have +1 oxidation state.
Question 29.
Why first ionization enthalpy of chromium is lower than that of zinc?
Answer:
The first ionization enthalpy of chromium is lower than that of zinc. $\mathrm{Cr}(\mathrm{Z}=24)$ Electronic configuration [Ar] $3 d^5 4 s^1$. In the case of $\mathrm{Cr}$, first electron has to be removed easily from 4 orbital to attain the more stable half filled configuration. So $\mathrm{Cr}$ has lower ionization enthalpy. But in the case of Zinc $(Z=30)$, electronic configuration [Ar] $3 \mathrm{~d}^{10} 4 \mathrm{~s}^2$. The first electron has to be removed from the most stable fully filled electronic configuration becomes difficult and it requires more energy.
Question 30.
Transition metals show high melting points why?
Answer:

1. All the transition metals are hard.
2. Most of them are hexagonal close packed, cubic close packed (or) body centered cubic which are characteristics of true metals.
3. The maximum melting point at about the middle of transition metal series indicates that $d^5$ configuration of favourable for strong inter atomic attraction.
4. Due to the strong metallic bonds, atoms of the transition elements are closely packed and held together. This leads to high melting point and boiling point.
Question 1.
Compare the stability of $\mathrm{Ni}^{4+}$ and $\mathrm{Pt}^{4+}$ from their ionisation enthalpy values
Answer:

$\mathrm{Pt}^{4+}$ compounds are stable than $\mathrm{Ni}^{4+}$ compounds because the energy needed to remove 4 electrons in $\mathrm{Pt}$ is less than that of $\mathrm{Ni}$.
Question 2.
Why iron is more stable in +3 oxidation state than in +2 and the reverse is true for Manganese?
Answer:
$\mathrm{Fe}(\mathrm{Z}=26)$. Electronic configuration $[\mathrm{Ar}] 3 \mathrm{~d}^6 4 \mathrm{~s}^2$
$\mathrm{Fe} \rightarrow \mathrm{Fe}^{3+}+3 \mathrm{e}^{-}$
$\mathrm{Fe}^{3+}$ Electronic configuration $\mathrm{p}^{\circ}[\mathrm{Ar}] 3 \mathrm{~d}^5$.
If ' $\mathrm{d}^{\prime}$ orbital is half filled, it is more stable than. $\mathrm{Fe}^{2+}$ where it is $[\mathrm{Ar}] 3 \mathrm{~d}^6$.
$\mathrm{Mn}(\mathrm{Z}=25)$. Electronic configuration $[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^2$
$\mathrm{Mn} \rightarrow \mathrm{Mn}^{2+}+2 \mathrm{e}^{-}$. By the loss of $2 \mathrm{e}^{-}, \mathrm{Mn}^{2+}$ is more stable due to half filled configuration. $\mathrm{Mn} \rightarrow \mathrm{Mn}^{3+}+$ $3 \mathrm{e}^{-}$
$\mathrm{Mn}^{3+}$ Electronic configuration $[\mathrm{Ar}] 3 \mathrm{~d}^4 4 \mathrm{~s}^{\circ}$.
Among this $\mathrm{Fe}^{3+}$ is more stable than $\mathrm{Fe}^{2+}$ and the $\mathrm{Mn}^{2+}$ is more stable than $\mathrm{Mn}^{3+}$.