InterviewSolution
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InterviewSolution
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The wrong statement regarding transition metals among the following is:A. `4s` electrons penetrate toward the nucleus more than `3d` electronsB. atomic radii of transition metals increase rapidly with increase in atomic number because of poor shielding of nuclear attraction by `(n-1)` of electronsC. second and third transition series elements have nearly the same sizeD. their densities are higher and densities of the `5d` series element are higher than those of `4d` series elements. |
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Answer» Correct Answer - B (A) The order of penetration of the electrons present in different sub-shells of same energy level is sgtpgtdgtf. (B) The decrease in size is small after mid way. In the beginning , the atomic radius decrease with the increase in nuclear charge (as atomic number increase), Whereas the shielding effect of d-electrons is small. After mid way as the electrons enter the last but one shell. the added d-electron shields the outer most. electrons . Hence with the increase in the d-electrons screening effect increase . this counter balances the increased nuclear charge . As a result , the atomic radii remain pratically same after chromium . (c ) The filling of `4f` before `5d` orbital results in a regular decrease in atomic radii called. Lanthanoids contraction which essentially compensates for the excepted increase in atomic size with increasing atomic number . The net result of the lanthanoid contraction is the seconf and the third d series exhibit similar radii(e.g., `Zr 160 pm, Hf 159 pm` ) (D) In transition elements , the atomic volumes are large because the increase nuclear charge is poorly screened and so attracts all the electron more strongly . In addition , the extra electrons added occupy inner orbitals. Consequently the densities of the transition metal are high . the densities of the second row are high and third row value higher because of lanthanoid contraction. |
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