The electronic configuration of four elements are:
(I) `[Kr]5s^(1)`
(II) `[Rn]5f^(14)6d^(1)7s^(2)`
(III) `[Ar]3d^(10)4s^(2)4p^(5)`
(IV) `[Ar]3d^(6)4s^(2)`
Consider the following statements:
(I I shows variable oxidation state
(ii) II is a d-block element
(iii) The compound formed between I and III is covalent
(iv) IV shows single oxidation state
which statement is true (T) or false (F)?

Options

A. FTFF
B. FTFT
C. FFTF
D. FFFF

Answer

D

Explanation

(I) `[Kr]5s^(1)`, shows only single oxidation state +1
(II) `[Rn]5f^(14)6d^(1)7s^(2)`, it is f-block element (Z=103)
(III) The compound formed between I and III is ionic.
(IV) `[Ar]3d^(6)4s^(2),` (Z=26) Fe shows variable oxidation state.