Question

If the second I.P of calcium is $275.12$ kcals $mo{l}^{-1}$, then, the equation that represents this ionization can be written as:

• A. $C{a}_{\left(S\right)}^{+}\to C{a}_{\left(S\right)}^{3+}+2e^{-}$ -275.12 kcals
• B. $C{a}_{\left(S\right)}^{+}\to C{a}_{\left(S\right)}^{3+}+2e^{-}$ +275.12 kcals
• C. $C{a}_{\left(S\right)}^{+}\to C{a}_{\left(S\right)}^{2+}+e^{-}$ +275.12 kcals
• D. $C{a}_{\left(S\right)}^{+}\to C{a}_{\left(S\right)}^{2+}+e^{-}$ -275.12 kcals

Answer 1

Answer: (D)

$C{a}_{\left(S\right)}^{+}\to C{a}_{\left(S\right)}^{2+}+e^{-}$ -275.12 kcals

I.P is the potential at which electron is removed from the element . Thus second ionization potential involves removal of second electron . Therefore the equation is $C{a}^{+}\to C{a}^{2+}+e^{-}$ $-275.12$ Kcals /mole