One element has atomic weight $39$. Its electronic configuration is $1{\mathrm{s}}^2,2{\mathrm{s}}^{2}2{\mathrm{p}}^6,3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^1.$ The true statement for that element is-

The ground state electronic configuration of elements $\mathrm{U}, \mathrm{V}, \mathrm{W}, \mathrm{X}$ and $\mathrm{Y}$ (these symbols do not have any chemical significance) are as follows:

$\mathrm{U}: 1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^3$

$\mathrm{V}: 1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^1$

$\mathrm{W}: 1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^2$

$\mathrm{X}: 1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{5}4{\mathrm{s}}^2$

$\mathrm{Y}: 1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{10}4{\mathrm{s}}^{2}4{\mathrm{p}}^6$

Determine which sequence of elements satisfy the following statements?

An element forms a carbonate which is not decomposed by heating.

An element is most likely to form coloured ionic compounds.

An element has a largest atomic radius.

Element forms only acidic oxides.

Consider the following change.

$\mathrm{M}\left(\mathrm{s}\right)\to \mathrm{M}\left(\mathrm{g}\right)$.... $\left(1\right)$

$\mathrm{M}\left(\mathrm{s}\right)\to {\mathrm{M}}^{2+}\left(\mathrm{g}\right)+2{\mathrm{e}}^{-}$....$\left(2\right)$

$\mathrm{M}\left(\mathrm{g}\right)\to {\mathrm{M}}^{+}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}$.... $\left(3\right)$

${\mathrm{M}}^{+}\left(\mathrm{g}\right)\to {\mathrm{M}}^{2+}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}$....$\left(4\right)$

$\mathrm{M}\left(\mathrm{g}\right)\to {\mathrm{M}}^{2+}\left(\mathrm{g}\right)+2{\mathrm{e}}^{-}$... $\left(5\right)$

The second ionization energy of $\mathrm{M}$ could be calculated from the energy values associated with: