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Consider the following species:
N3-, O2-, F, Na+, Mg2+, Al3+
(a) What is common in them?
(b) Arrange them in order of increasing ionic radii?

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Na with atomic no. 7 after gaining 3 electrons (Na³-)has total 10 (7+3) electrons. O with atomic no. 8 after gaining 2 electrons (O²-) has total 10 (8+2) electrons. F with atomic no. 9 after gaining 1 electron (F-) has total 10 (9+1) electrons. Na with atomic no. 11 after losing 1 electron...
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Na with atomic no. 7 after gaining 3 electrons (Na³-)has total 10 (7+3) electrons. 

O with atomic no. 8 after gaining 2 electrons (O²-) has total 10 (8+2) electrons. 

F with atomic no. 9 after gaining 1 electron (F-) has total 10 (9+1) electrons. 

Na with atomic no. 11 after losing 1 electron (Na+)has total 10 (11-1) electrons. 

Mg with atomic no. 12 after losing 2 electrons has total 10 (12-2) electrons. 

Al with atomic no. 13 after losing 3 electrons has total 10 (13-3) electrons. 

So these are all isoelectric species with equal number of electrons. 

b) Increasing order is Al³+, Mg²+, Na²+, F-, O²-, Na³-

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Tutor

i)Each of the given species (ions) has the same number of electrons (10 electrons). Hence, the given are isoelectronic species.N3- has 7+3 = 10 electronsO2- has 8+2= 10 electronsF- has 9+1 = 10 electronsNa+ has 11-1 = 10 electronsMg2+ has 12-2 = 10 electronsAl3+ has 13-3= 10 electronsii)Al3+ < Mg2+...
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i)Each of the given species (ions) has the same number of electrons (10 electrons). Hence, the given are isoelectronic species.
N3- has 7+3 = 10 electrons
O2- has 8+2= 10 electrons
F- has 9+1 = 10 electrons
Na+ has 11-1 = 10 electrons
Mg2+ has 12-2 = 10 electrons
Al3+ has 13-3= 10 electrons
ii)Al3+ < Mg2+ < Na+ < F– < O2– < N3–

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