Question

Find the value of P from rhe following properties. 7:P::35:45​

Answer 1

Answer:

$\begin{gathered}\Large{\underline{\underline{\bf{\color{peru}Iso\:Electronic\:Species\:;-}}}} \\ \end{gathered}$IsoElectronicSpecies;−

☆ Those having same number of electrons \bf{(e^-)}(e−) but different nuclear charge forms iso-electronic series.

$\begin{gathered}\bf\red{Examples,} \\ \end{gathered}$Examples,

1) $\bf{O^{2-}\:,\:F^{-}\:,\:Mg^{2+}}$O2−,F−,Mg2+ have 10 electrons. Hence, they are iso-electronic species.

2) $\bf{K^{+}\:,\:Ca^{2+}\:,\:S^{2-}\:,\:Cl^-}$K+,Ca2+,S2−,Cl− have 18 electrons. Hence, they are iso-electronic species.

3) $\bf{Ne\:,\:F^{-}}$Ne,F− have 10 electrons. Hence, they are iso-electronics.

4) $\bf{N^{3-}\:,\:Al^{3+}\:,\:Mg^{2+}}$N3−,Al3+,Mg2+ have 10 electrons. Hence, they are iso-electronics.

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♕ Now we take one the above example and clarify it.

✒ We take second example.

$\begin{gathered}\boxed{\begin{array}{cccc}\bf Species & \bf Z & \bf e^- \\ \frac{\qquad \qquad \qquad \qquad}{} & \frac{\qquad \qquad \qquad \qquad}{} & \frac{\qquad \qquad \qquad \qquad\qquad}{} \\ \bf K^+ & \sf 19 & \sf 18 \\ \\ \bf Ca^{2+} & \sf 20 & \sf 18 \\ \\ \bf S^{2-} & \sf 16 & \sf 18 \\ \\ \bf Cl^- & \sf 17 & \sf 18 \end{array}}\end{gathered}$SpeciesK+Ca2+S2−Cl−Z19201617e−18181818

$\begin{gathered}\bf\blue{Where,} \\ \end{gathered}$

Z denotes the total number of electrons are present in that element.

$\bf{e^-}$e− denotes the number of electrons are present after gain/loss of valency electrons

Answer 2

Answer:

7:P::35:45, The true value of P is 9.