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Practice each question, then open the answer or worked solution.
Why does copper not replace hydrogen from acids?
Ans: Copper lies below hydrogen in electrochemical series which means it has positive standard reduction potential i.e. \[{\text{ + 0}}{\text{.34 V}}\]. and hence cannot liberate hydrogen from acids.
Why \[{E^\circ }\] value for \[Mn\], \[Ni\] and \[Zn\] are more negative than expected?
Ans: \[{\text{Mn}}\] has half-filled 3d5 electrons and \[{\text{Zn}}\] has fully-filled 3d10 electrons which give them extra stability and they both resist to lose electrons and get reduced.
The \[{{\text{E}}^{{\circ }}}\] value also depends on the hydration enthalpy. More negative is the enthalpy of hydration, high is the \[{{\text{E}}^{{\circ }}}\] value. This is the reason behind\ [{\text{Ni}}\] having higher \[{{\text{E}}^{{\circ }}}\] value than expected.
Why is the first ionization enthalpy of \[Cr\] lower than that of \[Zn\]?
Ans: Cr(24)= [Ar]3d54s2
${{\text{Zn }}\left( {{\text{30}}} \right){\text{ = }}\left[ {{\text{Ar}}} \right]{\text{ 3}}{{\text{d}}^{\text{10}}}{\text{4}}{{\text{s}}^{\text{2}}}}$
The stability of orbitals according to the electrons filled is in order:
Fully filled > half-filled > partially filled
As \[{\text{3}}{{\text{d}}^{{\text{10}}}}\] is completely filled orbital of \[{\text{Zn}}\] is more stable than the \[{\text{3}}{{\text{d}}^{\text{5}}}\] half-filled orbital of \[{\text{Cr}}\] . It is comparatively easy to remove electron from \[{\text{3}}{{\text{d}}^{\text{5}}}\] than from \[{\text{3}}{{\text{d}}^{{\text{10}}}}\].
Therefore, the first ionization enthalpy of \[{\text{Cr}}\] is lower than that of \[{\text{Zn}}\].
Transition elements show high melting points. Why?
Ans: Transition elements have the ability to involve their d-electrons along with their s-electrons in the process of bond formation. This results in formation of strong metal-metal bonds which makes their melting point high.
When \[C{u^{2 + }}\] ion is treated with \[{\text{KI}}\], a white precipitate is formed. Explain the reaction with the help of a chemical equation.
Ans: ${\text{2C}}{{\text{u}}^{{\text{2 + }}}}{\text{ + 4}}{{\text{I}}^{\text{ - }}}{\text{ }} \to {\text{ C}}{{\text{u}}_{\text{2}}}{{\text{I}}_{\text{2}}}{\text{ + }}{{\text{I}}_{\text{2}}}$
When \[{\text{C}}{{\text{u}}^{{\text{2 + }}}}{\text{ }}\]reacts with potassium iodide white precipitate of \[{\text{C}}{{\text{u}}_{\text{2}}}{{\text{I}}_{\text{2}}}{\text{ }}\]is formed.
Out of \[C{u_2}C{l_2}\] and \[CuC{l_2}\], which is more stable and why?
Ans: Higher enthalpy of hydration provides extra stability to \[{\text{C}}{{\text{u}}^{{\text{2 + }}}}\] oxidation state than +1 oxidation state.
Enthalpy of hydration is described as the amount of energy released on dilution of one mole of gaseous ions.
When a brown compound of manganese (A) is treated with \[HCl\], it gives a gas
(B) The gas taken in excess, reacts with \[N{H_3}\] to give an explosive compound
(C). Identify compounds A, B and C.
Ans: The reactions are explained as:
${\text{Mn}}{{\text{O}}_{\text{2}}}{\text{ + 4HCl }} \to {\text{ MnC}}{{\text{l}}_{\text{2}}}{\text{ + C}}{{\text{l}}_{\text{2}}}{\text{ + 2}}{{\text{H}}_{\text{2}}}{\text{O}}$
${\text{ (A) (B)}}$
${\text{3C}}{{\text{l}}_{\text{2}}}{\text{ + 2N}}{{\text{H}}_{\text{3}}}{\text{ }} \to {\text{ NC}}{{\text{l}}_{\text{3}}}{\text{ + 3HCl}}$
${\text{(excess) (C)}}$
Although fluorine is more electronegative than oxygen, the ability of oxygen to stabilize higher oxidation states exceeds that of fluorine. Why?
Ans: Fluorine - 1s22s22p5
${{\text{Oxygen - 1}}{{\text{s}}^{\text{2}}}{\text{ 2}}{{\text{s}}^{\text{2}}}{\text{ 2}}{{\text{p}}^{\text{6}}}{\text{ 3}}{{\text{s}}^{\text{2}}}{\text{ 3}}{{\text{p}}^{\text{6}}}} $
As we can infer from the electronic configuration, fluorine only has one unpaired electron in its 2p-orbital which can be involved in formation of only one bond and fluorine lacks d-orbital.
While oxygen has d-orbitals in its configuration which allow the oxygen to form multiple bonds. Therefore, oxygen has greater stability to stabilise its oxidation states than fluorine.
Although \[C{r^{3 + }}\] and \[C{o^{2 + }}\] ions have same number of unpaired electrons but the magnetic moment of Cr3+ is 3.87 BM and that of \[C{o^{2 + }}\] is 4.87 BM. Why?
Ans: Magnetic moment of an electron/dipole moment is caused by its intrinsic properties of spin and electric charge. It depends upon the number of unpaired electrons in its valence shell. The more the number of unpaired electrons, the greater will be the value of magnetic moment.
${\text{C}}{{\text{r}}^{{\text{3 + }}}}{\text{ = }}\left[ {{\text{Ar}}} \right]{\text{ 3}}{{\text{d}}^{\text{3}}}{\text{ = 3 unpaired electrons}}$
${\text{C}}{{\text{o}}^{{\text{2 + }}}}{\text{ = }}\left[ {{\text{Ar}}} \right]{\text{ 3}}{{\text{d}}^7}{\text{ = 3 unpaired electrons}}$
But as both the ions have the same no. of unpaired electrons, there’s one other factor called orbital contribution which also influences the magnetic moment. As, the three unpaired electronic configuration is symmetrical in \[{\text{C}}{{\text{r}}^{{\text{3 + }}}}\] this contributes zero orbital contribution but this is quite appreciable in case of \[{\text{C}}{{\text{o}}^{{\text{2 + }}}}\] and hence it has larger value of Magnetic moment than \[{\text{C}}{{\text{r}}^{{\text{3 + }}}}\]
Ionisation enthalpies of Ce, Pr and Nd are higher than Th, Pa and U. Why?
Ans: \[{\text{Ce, Pr and Nd}}\] belong to the lanthanide series whereas \[{\text{Th, Pa and U}}\] belong to the actinides family.
When electrons start accommodating the 4f and 5f orbitals, the 5f electrons penetrate less into the inner core. They are more effectively shielded nuclei in comparison to 4f-electrons in lanthanides. This leads to the fact that 5f-electrons experience reduced nuclear force of attraction and hence they have Lower ionisation enthalpies than lanthanoids.