Which one will show variable oxidation state?

(I) $\left[Kr\right]5s^1$                                 (II) $\left[Rn\right]5f^{14}6d^{1}7s^2$
(III)$\left[Ar\right]3d^{10}4s^{2}4p^5$                    (IV) $\left[Ar\right]3d^{6}4s^2$

In alkaline earth metals, the properties, from the following, that will increase from Be to Ba are-

(i) Atomic radius       
(ii) Ionisation energy       
(iii) Nuclear charge

Incorrect order of ionic size is :

(a)$L{a}^{3+}>G{d}^{3+}>E{u}^{3+}>L{u}^{3+}$           (b)$V^{2+}>V^{3+}>V^{4+}>V^{5+}$

(c)$T{l}^{+}>I{n}^{+}>S{n}^{2+}>S{b}^{3+}$                (d)$K^{+}>S{c}^{3+}>V^{5+}>M{n}^{7+}$

Consider the following changes:

 $M\left(s\right)\to M\left(g\right) .....\left(i\right)$
$M\left(s\right)\to M^{2+}\left(g\right)+2e^{- } .....\left(ii\right)$
$M\left(g\right)\to M^{+}\left(g\right)+e^{-} .....\left(iiii\right)$
$M^{+}\left(g\right)\to M^{2+}\left(g\right)+e^{- } ......\left(iv\right)$
$M\left(g\right)\to M^{2+}\left(g\right)+2e^{-} ......\left(v\right)$

The second ionization energy of M could be calculated from the energy values associated with :

(a) 1+3+4                  (b) 2-1+3

(c) 1+5                      (d) 5-3

Which of the following statements is incorrect?

Aqueous solutions of two compounds $M_1-O-H and M_2-O-H$ are prepared in two different beakers. If, the electronegativity of $M_1=3.4 , M_2=1.2, O=3.5 and H=2.1,$ then the nature of two solutions will be respectively:

(a) acidic, basic             (b) acidic, acidic

(c) basic, acidic             (d) basic, basic

First three ionisation energies (in kJ/mol) of three representative elements are given below:

Element     $I{E}_1$                    $I{E}_2$                   $I{E}_3$

P               495.8                4562                  6910
Q               737.7               1451                   7733
R               577.5                1817                  2745

Then incorrect option is :

1. Q: Alkaline earth metal.

2. P: Alkali metal.

3. R: s-block element.

4. All three: P,Q & R belong to the same period.

If the ionization enthalpy and negative electron gain enthalpy of an element are 275 and 86 kcal $mo{l}^{-1}$ respectively, then the electronegativity of the element on the Pauling scale is:

1. 2.8                                     

2. 0.0

3. 4.0                                     

4. 2.6

The incorrect statement is:

(1) The second ionisation energy of Se is greater than that of second ionisation energy of As 

(2) The first ionisation energy of $C^{2+}$ ion is greater than that of first ionisation energy of $N^{2+}$ ion

(3) The third ionisation energy of F is greater than that of third ionisation energy of O

(4) Helogens have highest I.E. in respective period

Consider the following ionisation reactions :

I.E. (kJ $mo{l}^{-1}$)

$A_{\left(g\right)}\to A_{\left(g\right)}^{+}+e^{-} , A_1$
$B_{\left(g\right)}^{+}\to B_{\left(g\right)}^{2+}+e^{-}, B_2$
$C_{\left(g\right)}^{+}\to C_{\left(g\right)}^{2+}+e^{-}, C_2$
$B_{\left(g\right)}\to B_{\left(g\right)}^{+}+e^{-}, B_1$
$C_{\left(g\right)}\to C_{\left(g\right)}^{+}+e^{-}, C_1$
$C_{\left(g\right)}^{2+}\to C_{\left(g\right)}^{3+}+e^{-}, C_3$

If monovalent positive ion of A, divalent positive ion of B and trivalent positive ion of C have zero electron. Then incorrect order of corresponding I.E. is :

(1) $C_3>C_2>B_2$                       

(2) $B_1>A_1>C_1$

(3) $C_3>C_2>B_2$                       

(4) $B_2>C_3>A_1$