D and F Block Elements MCQ Questions & Answers in Inorganic Chemistry | Chemistry

When the transition metals are in their highest oxidation state, they no longer have tendency to give away electrons, thus they are not basic but show acidic character and form anionic complexes.

The correct order of ionic radii of $${Y^{3 + }},L{a^{3 + }},E{u^{3 + }}$$   and $$L{u^{3 + }},$$  is $${Y^{3 + }} < L{u^{3 + }} < E{u^{3 + }} < L{a^{3 + }}$$       because $$Eu$$  and $$Lu$$  are the members of lanthanide series (so they show lanthanide contraction) and $$La$$  is the representative element of all elements of such series and $${Y^{3 + }}$$  $$ion$$  has lower radii as comparison to $$L{a^{3 + }}$$  because it lies immediately above it in the periodic table.

$$\left( 1 \right)\,\,V = 3{d^3}4{s^2}\,\,\,\,\,\,\,\,{V^{2 + }} = 3{d^3} = $$       3 unpaired electron
$$\,\,\,\,\,\,\,\,\,Cr = 3{d^5}4{s^1}\,\,\,\,\,\,C{r^{2 + }} = 3{d^4} = $$       4 unpaired electron
$$\,\,\,\,\,\,\,\,Mn = 3{d^5}4{s^2}\,\,\,\,M{n^{2 + }} = 3{d^5} = $$       5 unpaired electron
$$\,\,\,\,\,\,\,Fe = 3{d^6}4{s^2}\,\,\,\,\,\,F{e^{2 + }} = 3{d^6} = $$       4 unpaired electron
hence the correct order of paramagnetic behaviour
$${V^{2 + }} < C{r^{2 + }} = F{e^{2 + }} < M{n^{2 + }}$$
$$\left( 2 \right)$$  For the same oxidation state, the ionic radii generally decreases as the atomic number increases in a particular transition series. hence the order is
$$M{n^{ + + }} > F{e^{ + + }} > C{o^{ + + }} > N{i^{ + + }}$$
$$\left( 3 \right)$$  In solution, the stability of the compounds depends upon electrode potentials, SEP of the transitions metal ions are given as
$$C{o^{3 + }}/Co = + 1.97,F{e^{3 + }}/Fe = + 0.77;$$        $$C{r^{3 + }}/C{r^{2 + }} = - 0.41,S{c^{3 + }}$$     is highly stable as it does not show $$ + 2\,O.S.$$
$$\eqalign{ & \left( 4 \right)Sc - \left( { + 2} \right),\left( { + 3} \right) \cr & \,\,\,\,\,\,Ti - \left( { + 2} \right),\left( { + 3} \right),\left( { + 4} \right) \cr & \,\,\,\,\,\,Cr - \left( { + 1} \right),\left( { + 2} \right),\left( { + 3} \right),\left( { + 4} \right),\left( { + 5} \right),\left( { + 6} \right) \cr & \,\,\,\,\,Mn - \left( { + 2} \right),\left( { + 3} \right),\left( { + 4} \right),\left( { + 5} \right),\left( { + 6} \right),\left( { + 7} \right) \cr} $$
i.e. $$Sc < Ti < Cr = Mn$$
$${E_a} = 53598.6\,{\text{J/mol}} = 53.6\,{\text{kJ/mol}}{\text{.}}$$

$$C{r^{3 + }} = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^3}$$       (coloured)
($$C{r^{3 + }}$$  contains 3 unpaired $${e^ - },$$  so it gives colour)
$$Z{n^{2 + }} = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}}$$       (colourless)
$$C{u^ + } = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}}$$       (colourless)
$$T{i^{4 + }} = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}$$       (colourless)
( Colour is produced due to presence of unpaired electrons. )

A - ii, B - i, C - iv, D - iii

$$S{O_2}$$  gas can readily oxidise acidified $$KMn{O_4}$$  solution because $$KMn{O_4}$$  is an oxidising agent and $$S{O_2}$$  act as reducing agent.
$$2MnO_4^ - + 5S{O_2} + 2{H_2}O \to $$      $$2M{n^{2 + }} + 5SO_4^{2 - } + 4{H^ + }$$
While other options such as $$N{O_2}$$  ( strong oxidising agent ), $$C{O_2}$$  ( neither oxidising agent nor reducing agent ) cannot decolourise acidified $$KMn{O_4}$$  Solution.

Change in $$O.S.$$  is by 1. Hence $$eq.\,wt.$$  is $$\frac{{158}}{1} = 158.$$

In $$KI{O_4},O.S$$   of $$I$$  is +7 and in $$KMn{O_4},O.S.$$    of $$Mn$$  is +7.

Due to presence of unpaired electrons in $$\left( {n - 1} \right)$$  $$d$$ - orbitals, the most of the transition metal ions and their compounds are paramagnetic. They form coloured ions and show variable oxidation states due to presence of vacant $$d$$ - orbitals.

More the number of unpaired $$d$$ - electrons, more is the magnetic moment.
$$C{u^{2 + }} - 3{d^9}$$     No. of unpaired electrons = 1
$$N{i^{2 + }} - 3{d^8}$$     No. of unpaired electrons = 2
$$C{o^{2 + }} - 3{d^7}$$     No. of unpaired electrons = 3
$$F{e^{2 + }} - 3{d^6}$$     No. of unpaired electrons = 4