The size of isoelectronic species F, Ne and Na+ is affected by

(a) nuclear charge (Z )

(b) valence principal quantum number (n)

(c) electron-electron interaction in the outer orbitals

(d) none of the factors because their size is the same.

Asked by Pragya Singh | 1 year ago |  193

Solution :-

Right answer is (a) Nuclear charge (Z)

Explanation:-

Because for isoelectronic species:

The atomic size decreases with an increase in nuclear charge (Z).

e.g. the arrangement according to increasing order of nuclear charge (Z) for

$$F^{-}, Ne\;and\;Na^{+} is$$

$$F^{-} < Ne < Na^{+}$$

And the arrangement according to increasing order of atomic size for

$$F^{-}, Ne\;and\;Na^{+} is$$

$$Na^{+} < Ne < F^{-}$$

Answered by Abhisek | 1 year ago

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