Q.The size of isoelectronic species — , Ne and is affected by
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Start your 14-day free trial to unlock the full solution →For isoelectronic species, the ionic/atomic radius is determined solely by the effective nuclear charge () — the higher the nuclear charge, the more strongly the electrons are pulled in, making the radius smaller. Among , Ne, and , the order of decreasing size is , so option (a) is correct.
Why effective nuclear charge is the key
When two or more species have the same number of electrons (isoelectronic), the electron-electron repulsion is identical in each case — the same number of electrons occupy the same set of orbitals. The only thing that changes is the nuclear charge (the number of protons). A higher pulls the electron cloud inward more strongly, shrinking the radius. The valence principal quantum number is the same for all three (they all have electrons in the shell), so it cannot explain the size difference. Electron-electron interaction is also identical because the electron count is the same. Therefore, the deciding factor is nuclear charge.
For isoelectronic species, radius , where (Slater’s rules). Since is nearly constant for the same electron configuration, increases with .
Step-by-step reasoning
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Identify the electron configuration.
has 9 protons and 10 electrons. Ne has 10 protons and 10 electrons. has 11 protons and 10 electrons. All three have the same electron configuration: (the neon core). They are isoelectronic.
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Compare nuclear charges.
- :
- Ne:
- :
The number of electrons is fixed at 10, so the shielding constant (from the 1s and 2s/2p electrons) is essentially the same for all three. Hence increases directly with .
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Relate to size.
A higher means a stronger pull on the valence electrons, reducing the atomic/ionic radius. So the species with the smallest (least pull) will be the largest, and the one with the largest (strongest pull) will be the smallest.
- (): lowest → largest radius
- Ne (): intermediate → intermediate radius
- (): highest → smallest radius …
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