NEET Practice Questions (MCQs) with Answers & Solutions

K₂Cr₂O₇ + 3Na₂SO₃ + 4H₂SO₄ $\to$3Na₂SO₄ + K₂SO₄ + Cr₂(SO₄)₃ + 4H₂O

Key Idea Colour is obtained as a consequence of d-d (or f-f) d-d (or M transition. and for d-d (or f-f) transition, presence of unpaired electrons is the necessary condition.

La³⁺(Z=57)=[Xe] 4f⁰5d⁰6s⁰ (no unpaired electron)

Ti³⁺(Z=22)=[Ar] 3d¹4s⁰ (one unpaired electron)

Lu³⁺(Z=71)=[Xe]4f¹⁴5d⁰6s⁰ (no unpaired electron)

Sc³⁺(Z=21)=[Ar] 3d⁰ 4s⁰ (no unpaired electron)

Hence, only Ti³⁺ will exhibit colour in aqueous solution.

(c) The regular decrease in the radii of lanthanide ions from La3+ to Lu3+ is known as lanthanide contraction. It is due to the greater effect of the increased nuclear charge than that of the screening effect (shield effect). As a result of lanthanide contraction, the atomic radii of element of 4d and 5d one close just above them in their respective group, so the properties of 4d and 5d transition element shows the similarities.

(b) More number of oxidation states are exhibited by the acinoids than by the lanthanoids due to lesser energy difference between 5f and 6d orbitals than that between 4f and 5d orbitals.

(a) Wilkinson's catalyst is $\left[\mathrm{RhCl}\right({\mathrm{PPh}}_3{)}_3]$, Red-violet in colour and has square planar structure. It is used for selective hydrogenation of organic molecules at room temperature and pressure.

${\mathrm{TiCl}}_4+({\mathrm{C}}_2{\mathrm{H}}_5{)}_3\mathrm{Al}$ is Zeigler Natta catalyst.

$({\mathrm{C}}_2{\mathrm{H}}_5{)}_4\mathrm{Pb}$ is an anti-knocking agent.

cis-platin is used as an anti-cancer agent.

(b) 

$$\begin{gathered} \underset{\mathrm{Acid}}{\mathrm{Zn}(\mathrm{OH}{)}_2}+\underset{\mathrm{Base}}{2{\mathrm{OH}}^{-}}\underset{}{\to } \underset{\mathrm{Salt}}{{\mathrm{ZnO}}_2^{2-}}+\underset{\mathrm{water}}{2{\mathrm{H}}_2\mathrm{O}} \\ \underset{\mathrm{Base}}{\mathrm{Zn}(\mathrm{OH}{)}_2}+2{\mathrm{H}}^{+}\to \underset{\mathrm{Salt}}{{\mathrm{Zn}}^{2+}}+\underset{\mathrm{water}}{2{\mathrm{H}}_2\mathrm{O}} \end{gathered}$$

(b) Interstitial compound ${\mathrm{Fe}}_3{\mathrm{C}}_{}$ (cementide) is formed.

(b) ${\mathrm{Na}}_2\left[\stackrel{\mathrm{II}}{\mathrm{Fe}}\right(\mathrm{NO}\left)\right(\mathrm{CN}{)}_2] + {\mathrm{Na}}_2\mathrm{S} \stackrel{}{\to }{\mathrm{Na}}_4[\stackrel{\mathrm{II}}{\mathrm{Fe}}\left(\mathrm{CN}{)}_5\right(\mathrm{NOS}\left)\right]$