Question

The ionic radii of N³- ,O²- and F-are respectively given by —
(1) 1.36, 1.40, 1.71
(2) 1.36, 1.71, 1.40
(3) 1.71, 1.40, 1.36
(4) 1.71, 1.36, 1.40

Answer 1

answer :1.36 ,1.40,1.71

option a is the answer

The ions F-, O2- and N3- are isoelectronic (same number of electrons) and belongs to the same period (Period 2). The size of such ions decreases with increase in atomic number or decreases moving from left to right or increases with the increase in the negative charge.

Answer 2

The ionic radii of $N^{3-}$ ,$O^{2-}$ and  $F^-$ are respectively 1.71 , 1.40 and 1.36

Explanation:

Isoelectronic species are defined as the molecules which have the same number of electrons.

Atomic number of nitrogen is 7 and thus has 7 electrons. $N^{3-}$ is formed by gain of 3  electrons and thus has 10 electrons.

Atomic number of oxygen is 8 and thus has 8 electrons. $O^{2-}$  is formed by gain of 2 electrons and thus has 10 electrons.

Atomic number of flourine is 9 and thus has 9 electrons. $F^-$ is formed by gain of 1 electron and thus has 10 electrons

As the number of electrons are same , the more is the nuclear charge i.e. the number of protons , more will the force of attraction towards the nuclei and thus smaller will be the radii and vice versa.Thus, order of increasing ionic radii for the given ions is as follows:

$N^{3-}$ > $O^{2-}$ > $F^-$

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