d and f Block Elements

NEET Chemistry · 90 questions · Page 2 of 9 · Click an option or "Show Solution" to reveal answer

Q11

Which of the following reactions is the metal displacement reaction? Choose the right option.

2Pb{(N{O_3})_2} \to 2PbO + 4N{O_2} + {O_2} \uparrow

2KCl{O_3}\overset{\Delta}\longrightarrow 2KCl + 3{O_2}

C{r_2}{O_3} + 2Al\overset{\Delta}\longrightarrow A{l_2}{O_3} + 2Cr

Fe + 2HCl \to FeC{l_2} + {H_2} \uparrow

Correct Answer

Option C

Solution

$\bullet$ Both reactions (a) and (b) are examples of decomposition reactions.

$\bullet$ Reactions (c) and (d), both are examples of displacement reactions, while reaction (c) is an example of metal displacement reaction n in which a more reactive metal displaces/replaces the less reactive metal.

Q12

Zr (Z = 40) and Hf (Z = 72) have similar atomic and ionic radii because of :

A Having similar chemical properties

B Belonging to same group

C Diagonal relationship

D Lanthanoid contraction

Correct Answer

Option D

Solution

$\bullet$ Hf is a post lanthanoid element.

Due to presence of4f-orbitals which have poor shielding effect, the effective nuclear charge on valence shell electrons is more which result in the decrease of the size of Hf.

This effect is known as lanthanoid contraction.

$\bullet$ The almost identical radii of Zr (160 pm) and Hf (159 pm) is a consequence of the lanthanoid contraction.

Q13

Identify the incorrect statement.

A The transition metals and their compounds are known for their catalytic activity due to their ability to adopt multiple oxidation states and to form complexes.

B Interstitial compounds are those that are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals.

C The oxidation states of chromium in CrO 4 2- and Cr 2 O 7 2- are not the same.

D Cr 2+ (d 4 ) is a stronger reducing agent than Fe 2+ (d 6 ) in water.

Correct Answer

Option C

Solution

Oxidation state of Cr in CrO 4 2- and Cr 2 O 7 2- is + 6.

Q14

Which one of the following ions exhibits d-d transition and paramagnetism as well?

A CrO 4 2-

B Cr 2 O 7 2–

C MnO 4 -

D MnO 4 2-

Correct Answer

Option D

Solution

In CrO 4 2- Cr +6 (n = 0) diamagnetic In Cr 2 O 7 2– Cr +6 (n = 0) diamagnetic In MnO 4 - Mn +7 (n = 0) diamagnetic In MnO 4 2- Mn +6 (n = 1) paramagnetic In MnO 4 2- , one unpaired electron(n) is present in d-orbital so, d-d transition is possible.

Q15

Match the metal ions given in Column I with the spin magnetic moments of the ions given in Column II and assign the correct code : <table class=tg> <thead> <tr> <th class=tg-7hap>Column - I</th> <th class=tg-7hap>Column - II</th> </tr> </thead> <tbody> <tr> <td class=tg-60hs>A. Co<sup>3+</sup></td> <td class=tg-60hs>(i)

\sqrt 8

B.M.</td> </tr> <tr> <td class=tg-60hs>B. Cr<sup>3+</sup></td> <td class=tg-60hs>(ii)

\sqrt {35}

B.M.</td> </tr> <tr> <td class=tg-60hs>C. Fe<sup>3+</sup></td> <td class=tg-60hs>(iii)

\sqrt 3

B.M.</td> </tr> <tr> <td class=tg-60hs>D. Ni<sup>2+</sup></td> <td class=tg-60hs>(iv)

\sqrt {24}

B.M.</td> </tr> <tr> <td class=tg-60hs></td> <td class=tg-60hs>(v)

\sqrt {15}

B.M.</td> </tr> </tbody> </table>

A <table class=tg> <thead> <tr> <th class=tg-7hap>A</th> <th class=tg-7hap>B</th> <th class=tg-7hap>C</th> <th class=tg-7hap>D</th> </tr> </thead> <tbody> <tr> <td class=tg-60hs>(iv)</td> <td class=tg-60hs>(v)</td> <td class=tg-60hs>(ii)</td> <td class=tg-60hs>(i)</td> </tr> </tbody> </table>

B <table class=tg> <thead> <tr> <th class=tg-7hap>A</th> <th class=tg-7hap>B</th> <th class=tg-7hap>C</th> <th class=tg-7hap>D</th> </tr> </thead> <tbody> <tr> <td class=tg-60hs>(i)</td> <td class=tg-60hs>(ii)</td> <td class=tg-60hs>(iii)</td> <td class=tg-60hs>(iv)</td> </tr> </tbody> </table>

C <table class=tg> <thead> <tr> <th class=tg-7hap>A</th> <th class=tg-7hap>B</th> <th class=tg-7hap>C</th> <th class=tg-7hap>D</th> </tr> </thead> <tbody> <tr> <td class=tg-60hs>(iv)</td> <td class=tg-60hs>(i)</td> <td class=tg-60hs>(ii)</td> <td class=tg-60hs>(iii)</td> </tr> </tbody> </table>

D <table class=tg> <thead> <tr> <th class=tg-7hap>A</th> <th class=tg-7hap>B</th> <th class=tg-7hap>C</th> <th class=tg-7hap>D</th> </tr> </thead> <tbody> <tr> <td class=tg-60hs>(iii)</td> <td class=tg-60hs>(v)</td> <td class=tg-60hs>(i)</td> <td class=tg-60hs>(ii)</td> </tr> </tbody> </table>

Correct Answer

Option A

Solution

Co 3+ = [Ar]3d 6 , unpaired e – (n) = 4 Spin magnetic moment ( $\mu$ ) =

\sqrt {4\left( {4 + 2} \right)} = \sqrt {24}

B.M Cr 3+ = [Ar]3d 3 , unpaired e – (n) = 3 Spin magnetic moment ( $\mu$ ) =

\sqrt {3\left( {3 + 2} \right)} = \sqrt {15}

B.M. Fe 3+ = [Ar]3d 5 , unpaired e – (n) = 5 Spin magnetic moment ( $\mu$ ) =

\sqrt {5\left( {5 + 2} \right)} = \sqrt {35}

B.M. Ni 2+ = [Ar]3d 8 , unpaired e – (n) = 2 Spin magnetic moment ( $\mu$ ) =

\sqrt {2\left( {2 + 2} \right)}

\sqrt 8

B.M.

Q16

HgCl 2 and I 2 both when dissolved in water containing I

-

ions, the pair of species formed is

A HgI 2 , I

-

B HgI

_4^{2 - }

, I

_3^{ - }

C Hg 2 I 2 , I

-

D HgI 2 , I

_3^{ - }

Correct Answer

Option B

Solution

HgCl 2 + 4I – (aq) $\to$ HgI 4 2– (aq) + 2Cl – (aq) I 2 (s) + I – (aq) $\to$ I 3 – (aq)

Q17

Name the gas that can readily decolourise acidified KMnO 4 solution.

A SO 2

B NO 2

C P 2 O 5

D CO 2

Correct Answer

Option A

Solution

Acidified KMnO 4 is a strong oxidizing agent thus, among the given option which readily undergoes oxidation with KMnO 4 will decolourise it.

CO 2 , NO 2 and P 2 O 5 are already in their highest oxidation state while SO 2 can further oxidize with KMnO 4 to give sulphate ions.

2MnO 4 – + 5SO 2 + 2H 2 O $\to$ 2Mn 2+ + 5SO 4 2– + 4H +

Q18

The reason for greater range of oxidation states in actinoids is attributed to

A actinoid contraction

B 5

f

, 6d and 7s levels having comparable energies

C 4

f

and 5d levels being close in energies

D the radioactive nature of actinoids.

Correct Answer

Option B

Solution

Minimum or comparable energy gap between 5f, 6d and 7s subshell makes electron excitation easier, hence there is a greater range of oxidation states in actinoids.

Q19

Which one of the following statements related to lanthanoids is incorrect?

A Europium shows + 2 oxidation state.

B The basicity decreases as the ionic radius decreases from Pr to Lu.

C All the lanthanons are much more reactive than aluminium.

D Ce(+4) solutions are widely used as oxidizing agent in volumetric analysis.

Correct Answer

Option C

Solution

The first few members of the lanthanoid series are quite reactive, almost like calcium.

However, with increasing atomic number, their behavior becomes similar to that of aluminium.

Q20

Which one of the following statements is correct when SO 2 is passed through acidified K 2 Cr 2 O 7 solution?

A SO 2 is reduced.

B Green Cr 2 (SO 4 ) 3 is formed.

C The solution turns blue.

D The solution is decolourised.

Correct Answer

Option B

Solution

K 2 Cr 2 O 7 + H 2 SO 4 + 3SO 2 $\to$ K 2 SO 4 + Cr 2 (SO 4 ) 3 (Green) + H 2 O

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