During electrolysis, copper is deposited at the cathode while oxygen is liberated at anode. The following reactions occur at the electrodes At anode $$2{H_2}O \to 4{H^ + } + {O_2} + 4{e^ - }$$ (Oxidation) At cathode $$C{u^{2 + }} + 2{e^ - } \to Cu\left( s \right)$$ (Reduction)
112.
$$KMn{O_4}$$ acts as an oxidising agent in alkaline medium. When alkaline $$KMn{O_4}$$ is treated with $$KI,$$ iodide ion is oxidised to
In lanthanides, there is poorer shielding of $$5d$$ electrons by $$4f$$ electrons resulting in greater attraction of the nucleus over $$5d$$ electrons and contraction of the atomic radii.
114.
Which of the following statement is not correct ?
A
$$La{\left( {OH} \right)_3}$$ is less basic than $$Lu{\left( {OH} \right)_3}.$$
B
In lanthanide series ionic radius of $$L{n^{3 + }}\,ions$$ decreases.
C
$$La$$ is actually an element of transition series rather than lanthanide series.
D
Atomic radii of $$Zr$$ and $$Hf$$ are same because of lanthanide contraction.
Answer :
$$La{\left( {OH} \right)_3}$$ is less basic than $$Lu{\left( {OH} \right)_3}.$$
$$La{\left( {OH} \right)_3}$$ is more basic than $$Lu{\left( {OH} \right)_3}.$$ This is because ionic size of $$L{a^{3 + }}\,ion$$ is more than$$L{u^{3 + }}\,ion$$ (Fajan's rule).
115.
$$Zr$$ and $$Hf$$ have almost equal atomic and ionic radii because of
No explanation is given for this question. Let's discuss the answer together.
116.
On the basis of data given below,
$$\eqalign{
& E_{\frac{{S{c^{3 + }}}}{{S{c^{2 + }}}}}^\Theta = - 0.37,E_{\frac{{M{n^{3 + }}}}{{M{n^{2 + }}}}}^\Theta = + 1.57 \cr
& E_{\frac{{C{r^{2 + }}}}{{Cr}}}^\Theta = - 0.90,E_{\frac{{C{u^{2 + }}}}{{Cu}}}^\Theta = 0.34 \cr} $$
Which of the following statements is incorrect ?
A
$$S{c^{3 + }}$$ has good stability due of $$\left[ {Ar} \right]3{d^0}4{s^0}$$ configuration.
B
$$M{n^{3 + }}$$ is more stable than $$M{n^{2 + }}.$$
C
$$C{r^{2 + }}$$ is reducing in nature.
D
Copper does not give $${H_2}$$ on reaction with $$dil.\,{H_2}S{O_4}.$$
Answer :
$$M{n^{3 + }}$$ is more stable than $$M{n^{2 + }}.$$
The more extensive the metallic bonding of an element, the more will be its enthalpy of atomization. Due to the absence of unpaired electrons in zinc, metallic bonding is weakest and as a result, it has least enthalpy of atomization.
119.
Which of the following compounds has colour but no unpaired electrons ?
Electronic configuration of $$Mn$$ is $$\left[ {Ar} \right]3{d^5}4{s^2}.$$ Being transition metal it has 7 valence electrons and all are involved in bond formation in $$MnO_4^ - .$$ Hence it has no unpaired electron.
120.
Which of the following statement is incorrect ?
A
Across a period from $$Sc$$ to $$Cu$$ the densities increase with increasing atomic number.
B
The melting point of transition elements rise to a maximum from $$Sc$$ to $$Cr$$ and then decreases from $$Fe$$ to $$Zn.$$
C
Transition elements have high enthalpies of atomization and in $$3d$$ series increase regularly from $$Sc$$ to $$Cu.$$
D
On going down a group from $$3d$$ to $$6d$$ series the stability of higher oxidation state increases with increasing atomic number.
Answer :
Transition elements have high enthalpies of atomization and in $$3d$$ series increase regularly from $$Sc$$ to $$Cu.$$
(A), (B) and (D) are correct statements. (C) There is drop in enthalpy of atomization at manganese because of half filled stable electronic configuration leading to less mobilisation of electrons. This results in weaker metallic bonds.