d and f Block Elements

NEET Chemistry · 90 questions · Page 4 of 9 · Click an option or "Show Solution" to reveal answer

Q31

Which of the following lanthanoid ions is diamagnetic? (At. nos. Ce = 58, Sm = 62, Eu = 63, Yb = 70)

A Eu 2+

B Yb 2+

C Ce 2+

D Sm 2+

Correct Answer

Option B

Solution

Ion with no unpaired electron is diamagnetic in nature.

Ce 2+ = [Xe] 4f 2 → 2 unpaired electrons Sm 2+ = [Xe] 4f 6 → 6 unpaired electrons Eu 2+ = [Xe] 4f 7 → 7 unpaired electrons Yb 2+ = [Xe] 4f 14 → No unpaired electron

Q32

Which of the following statements about the interstitial compounds is incorrect?

A They are much harder than the pure metal.

B They have higher melting points than the pure metal.

C The retain metallic conductivity.

D They are chemically reactive.

Correct Answer

Option D

Solution

Interstitial compounds are generally chemically inert.

Q33

Which of the following exhibits only + 3 oxidation state?

A U

B Th

C Ac

D Pa

Correct Answer

Option C

Solution

U exhibits + 3, + 4, + 5, + 6 Th exhibits + 3, + 4 ; Ac exhibits + 3 only; Pa exhibits + 3, + 4, + 5

Q34

The catalytic activity of transition metals and their compounds is ascribed mainly to

A their magnetic behaviour

B their unfilled

d

-orbitals

C their ability to adopt variable oxidation states

D their chemical reactivity

Correct Answer

Option C

Solution

The transition metals and their compounds are used as catalysts because of the variable oxidation states.

Due to this, they easily absorb and re-emit wide range of energy to provide the necessary activation energy.

Q35

Four successive members of the first series of the transition metals are listed below. For which one of them the standard potential

\left( {E_{{M^{2 + }}/M}^o} \right)

value has a positive sign?

A Co (Z = 27)

B Ni (Z = 28)

C Cu (Z = 29)

D Fe (Z = 26)

Correct Answer

Option C

Solution

\left( {E_{{Co^{2 + }}/Co}^o} \right)

= -0.28

\left( {E_{{Ni^{2 + }}/Ni}^o} \right)

= -0.25

\left( {E_{{Cu^{2 + }}/Cu}^o} \right)

= +0.34

\left( {E_{{Fe^{2 + }}/Fe}^o} \right)

= -0.44

Q36

Which one of the following does not correctly represent the correct order of the property indicated against it?

A Ti < V < Cr < Mn; increasing number of oxidation states

B Ti 3+ < V 3+ < Cr 3+ < Mn 3+ : increasing magnetic moment

C Ti < V < Cr < Mn : increasing melting points

D Ti < V < Mn < Cr : increasing 2 nd ionization enthalpy

Correct Answer

Option C

Solution

The melting points of the transition elements first rise to a maximum and then fall as the atomic number increases, manganese have abnormally low melting points.

Q37

Identify the alloy containing a non-metal as a constituent in it.

A Invar

B Steel

C Bell metal

D Bronze

Correct Answer

Option B

Solution

Invar $\Rightarrow$ Ni(metal) + Fe(metal) Steel $\Rightarrow$ C(non-metal) + Fe(metal) Bell metal $\Rightarrow$ Cu(metal) + Sn(metal) + Fe(metal) Bronze $\Rightarrow$ Cu(metal) + Sn(metal)

Q38

Which of the statements is not true?

A On passing H 2 S through acidified K 2 Cr 2 O 7 solution, a milky colour is observed.

B Na 2 Cr 2 O 7 is preferred over K 2 Cr 2 O 7 in volumetric analysis.

C K 2 Cr 2 O 7 solution in acidic medium is orange.

D K 2 Cr 2 O 7 solution becomes yellow increasing the pH beyond 7.

Correct Answer

Option B

Solution

Na 2 Cr 2 O 7 is hygroscopic in nature and therefore accurate weighing is not possible in normal atmospheric conditions.

A hygroscopic substance absorbs moisture from atmosphere and this could lead to inaccuracies in weight.

Q39

Acidified K 2 Cr 2 O 7 solution turns green when Na 2 SO 3 is added to it. This is due to the formation of

A Cr 2 (SO 4 ) 3

B CrO

_4^{2 - }

C Cr 2 (SO 3 ) 3

D CrSO 4

Correct Answer

Option A

Solution

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

Q40

For the four successive transition elements (Cr, Mn, Fe and Co), the stability of + 2 oxidation state will be there in which of the following order? (At. nos. Cr = 24, Mn = 25, Fe = 26, Co = 27)

A Mn > Fe > Cr > Co

B Fe > Mn > Co > Cr

C Co > Mn > Fe > Cr

D Cr > Mn > Co > Fe

Correct Answer

Option A

Solution

The order can be explained using the idea of spin correlation.

Spin correlation refers to lowering of energy for like (parallel) spins.

Spin correlation leading to decrease in repulsion for electrons of like spins than for electrons of different spins is called exchange energy.

Spin correlation and its exchange energy gives an electronic configuration a special stability which is greatest for half-filled electronic configurations.

Mn 2+ (d 5 ) gets stabilisation due to half-filled configuration.

In Fe 2+ (d 6 ) the placing of one extra electron in a subshell destabilises.

Placing of 2 electrons in Co 2+ (d 7 ) destabilises it more.

Cr 2+ (d 4 ) has one vacant subshell.

Fe 2+ gets more stabilisation compared to Cr 2+ through exchange energy.

So the order is as follows: Mn > Fe > Cr > Co

Ready for a full NEET mock test? Timed · full syllabus · instant results

Take a Mock Test →