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11 Mar 2010 21:11:36 IST

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The first IE of carbon atom is greater than that of boron Whereas the reverse is true for the second

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The first IE of carbon atom is greater than that of boron Whereas the reverse is true for the second IE.

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Deepak Aggarwal

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12 Mar 2010 11:08:11 IST

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Configuration of C(6) = 1s² 2s² 2p²

and conf. of B(5) = 1s² 2s² 2p¹

Now first IE of C is greater as removal of one elctron is difficult bcoz it prefers to accept one e- rather. On the other hand after 1 e- is removed from each atom, 2nd e- is easier to be removed from C than B as boron attains highly stable conf. i.e. 1s² 2s² which is fully filled.

Remember : Difficult the removal, More the IE.

edison

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31 Jan 2011 16:10:34 IST

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The reason comes down to Coulomb's law (F = kq1q2/r^2) and the attraction of the negatively charged electrons for the positive charge of the nucleus AND the repulsion of the individual electrons. Let us pick the outermost electron. It is both attracted to the nucleus and repelled by each electron between it and the nucleus. This will produce a net force on the electron and govern in some measure the average distance between it and the nucleus.

In carbon, (1s2, 2s2, 2p2) and boron (1s2, 2s2, 2p1) there are unpaired outer electrons in the same energy level (2p). The repulsive forces of the inner electrons is essentially the same for carbon and boron. Carbon, with its greater positive nuclear charge has a slightly smaller atomic radius, and greater coulomic attraction for the outermost electron, hence the slightly greater first ionization energy.

When an outermost electron is removed carbon will have one electron in a p-sublelvel at roughly the same distance from the nucleus as the one that was removed. But for boron, the new outermost electron will be in a sublevel that is closer to the nucleus and therefore have greater attractive force according to Coulomb's law.

The combination of nuclear charge of the boron atom, and the smaller distance to the outermost electron, as well as the repulsion of the inner electrons, will produce a slightly greater force of attraction on the outermost electrons (in the 2s sublevel) of boron than for carbon. This results in a greater second ionization energy for boron.

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