The isolation of $\text{'}B\text{'}$ from $B_2{O}_3$ can be done by the following steps.

${\mathrm{B}}_2{\mathrm{O}}_3+\mathrm{M} \to \mathrm{Metal} \mathrm{oxide} + \mathrm{B} \underset{\& \mathrm{Filtered}}{\overset{\mathrm{dil}. {\mathrm{H}}_2{\mathrm{SO}}_4}{\to }} \mathrm{B}\left(\mathrm{s}\right) \mathrm{as} \mathrm{Residue}$

Which of the following metal is not suitable for the above process.

Boron is estimated from B₂O₃ using a metal that is electropositive and soluble in dilute sulfuric acid. The first step of the separation of B from B₂O₃ is done using metal and providing a large quantity. The second step is the purification of boron which is done using dilute H₂SO₄.

Using Na as a metal: Na is used for the estimation of Boron because it is electropositive and soluble in H₂SO₄.

${\mathrm{B}}_2{\mathrm{O}}_3+\mathrm{Na}\stackrel{∆}{\to }{\mathrm{Na}}_2\mathrm{O}+\underset{\mathrm{Impure}}{\mathrm{B}}\stackrel{\mathrm{dilute} {\mathrm{H}}_2{\mathrm{SO}}_4}{\to }\underset{\mathrm{Soluble}}{{\mathrm{Na}}_2{\mathrm{SO}}_4}+ \underset{\mathrm{Pure}}{\mathrm{B}}$

K is also used for the estimation of boron because it is electropositive and is soluble in dilute H₂SO₄.

${\mathrm{B}}_2{\mathrm{O}}_3+\mathrm{K}\stackrel{∆}{\to }{\mathrm{K}}_2\mathrm{O}+\underset{\mathrm{Impure}}{\mathrm{B}}\stackrel{\mathrm{dilute} {\mathrm{H}}_2{\mathrm{SO}}_4}{\to }\underset{\mathrm{Soulble}}{{\mathrm{K}}_2{\mathrm{SO}}_4}+\underset{\mathrm{Pure}}{\mathrm{B}}$

Ca is not used in the estimation because it is not soluble in dilute H₂SO₄, therefore pure boron cannot be achieved if Ca is used.

$\mathrm{Ca}+{\mathrm{B}}_2{\mathrm{O}}_3\stackrel{∆}{\to }\mathrm{CaO}+\underset{\mathrm{Impure}}{\mathrm{B}}\stackrel{\mathrm{dilute} {\mathrm{H}}_2{\mathrm{SO}}_4}{\to }\underset{\mathrm{Sparingly} \mathrm{soluble}}{{\mathrm{CaSO}}_4}+\underset{\mathrm{Impure} (\mathrm{as} \mathrm{it} \mathrm{have} \mathrm{traces} \mathrm{of} \mathrm{CaO})}{\mathrm{B}}$

Mg is also used in the estimation of boron because it is electropositive and is soluble in H₂SO₄.

${\mathrm{B}}_2{\mathrm{O}}_3+\mathrm{Mg}\stackrel{∆}{\to }\mathrm{MgO}+\underset{\mathrm{Impure}}{\mathrm{B}}\stackrel{\mathrm{dilute} {\mathrm{H}}_2{\mathrm{SO}}_4}{\to }\underset{\mathrm{Soluble}}{{\mathrm{MgSO}}_4}+\underset{\mathrm{Pure}}{\mathrm{B}}$