The electronic configuration of Palladium is:

1. [Rn] 5f³ 6d¹ 7s²

2. [Rn] 5f⁵ 6d¹ 7s²

3. [Rn] 5f² 6d¹ 7s²

4. None of the above.

The formation of the oxide ion O^2– (g), from the oxygen atom requires first an exothermic and then an endothermic step as shown below, 
$\mathrm{O}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}\to {\mathrm{O}}^{-};$ ${∆}_{\mathrm{f}}{\mathrm{H}}^0=–141$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
${\mathrm{O}}^{-}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}\to {\mathrm{O}}^{2-}\left(\mathrm{g}\right);$ ${∆}_{\mathrm{f}}{\mathrm{H}}^0=+780$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

Thus, the process of formation of O^2– in the gas phase is unfavorable even though O^2– is isoelectronic with neon. It is due to the fact that:

Identify the correct order of the size of the following:

1. $C{a}^{2+}<K^{+}<Ar<S^{2-}<C{l}^{-}$

2. $C{a}^{2+}<K^{+}<Ar<C{l}^{-}<S^{2-}$

3. $Ar<C{a}^{2+}<K^{+}<C{l}^{-}<S^{2-}$

4. $C{a}^{2+}<A{r}^{+}<K^{+}<C{l}^{-}<S^{2-}$

For the second-period elements, the correct increasing order of first ionisation enthalpy is:

1. Li < Be < B < C < O < N < F < Ne

2. Li < Be < B < C < N < O < F < Ne

3. Li < B < Be < C < O < N < F < Ne

4. Li < B < Be < C < N < O < F < Ne

The incorrect order of electronegativity is :

The element having electronic configuration [Kr]4d¹⁰4f¹⁴, 5s²5p⁶, 6s² belongs to:

1. s-block

2. p-block

3. d-block

4. f-block

The electronegativity of the following elements increases in the order:

1. C, N, Si, P

2. N, Si, C, P

3. Si, P, C, N

4. P, Si, N, C

The order of screening effect of electrons of s, p, d and f orbitals of a given shell of an atom on electrons in its outer shell is :

1. s > p > d > f

2. f > d > p > s

3. p < d < f < s

4. f > p > s > d

The elements in which electrons are progressively filled in 4f-orbitals are called: