Question

Variable valency is exhibited by:
A.Normal elements
B.Metallic elements
C.Transitional elements
D.Non metallic elements

Answer 1

Variable valency is shown by elements which have less energy difference between subshells of outermost shells. Metallic elements are generally elements which can donate electrons like s block elements. D block elements are also known as transition elements except ${\text{IIB}}$ elements. Elements of p block are non metallic elements.

Complete answer:
According to Bohr classification, Normal elements are those which have ${{\text{n}}^{{\text{th}}}}$ shell incomplete like s and p block elements except inert gas elements. Transition elements are those which have ${\text{ns}}$ and $\left( {{\text{n}} - 1} \right){\text{d}}$ shell incomplete either in atomic or ionic form of d block metals except ${\text{IIB}}$ elements.
Valency increases from ${\text{IA}}$ to ${\text{IVA}}$ and then decreases with respect to Hydrogen. On the other hand, valency increases from ${\text{IA}}$ to ${\text{VIIA}}$ with respect to Oxygen. Covalency is the sum of the number of covalent bonds and coordinate bonds formed by an atom. Electrovalency is the charge of ions. Maximum covalency of second period elements is 4. Third period elements can increase their covalency more than 4 due to the presence of vacant orbits. For s and p block elements, covalency increases by two with each excited state but for d block elements, covalency increases by one. According to Sidgwick concept, an atom which has vacant orbitals in its outermost shell can extend its covalency. For example: General covalency of Nitrogen is 3 but maximum covalency of Nitrogen is 4. The maximum oxidation state of Nitrogen is $+ 5$ . Oxidation state of p block elements lie in the range of $\left( {{\text{n}} - 8} \right){\text{to}}\left( { + {\text{n}}} \right)$ where n is the number of outermost electron of p block element.
Oxidation state of ${\text{IA}}$ is $+ 1$ and the oxidation state of ${\text{IIA}}$ is $+ 2$ . Generally, d block metals show variable oxidation state due to the less energy difference between subshell ${\text{ns}}$ and subshell $\left( {{\text{n}} - 1} \right){\text{d}}$ . Maximum oxidation state in the d block is $+ 8$ of Os and Ru. General and most stable oxidation state of Lanthanides is$+ 3$. Actinides show higher oxidation state than Lanthanides due to the less energy difference between 5f and 6d than that of 4f and 5d.

Thus, the correct option is C.

Note:
Oxygen has a strong tendency to stabilize the higher oxidation state of an element in comparison to Fluorine as Oxygen has a tendency of multiple bond formation.