It is not possible to measure the atomic radius precisely since the electron cloud surrounding the atom does not have a sharp boundary. One practical approach to estimate the size of an atom of a non-metallic element is to measure the distance between two atoms when they are bound together by a single bond in a covalent molecule and then dividing by two. For metals we define the term “metallic radius” which is taken as half the internuclear distance separating the metal cores in the metallic crystal. The van der waal’s radius represents the over all size of the atoms which includes its valence shell in a non bonded situation. It is the half of the distance between two similar atoms in separate molecules in a solid. The atomic radius decreases across a period and increases down the group. Same trends are observed in case of ionic radius. Ionic radius of the species having same number of electrons depends on the number of protons in their nuclei. Sometimes, atomic and ionic radii give unexpected trends due to poor shielding of nuclear charge by d- and f-orbital electrons.
Now answer the following three questions :
(i) Which of the following relations is correct, if considered for the same element :
Text Solution
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(i)
(ii) Ionic size ∝
for isoelectronic species.
(iii) Both N 3– and Al 3+ are isoelectronic species, but Al 3+ has greater nuclear charge. So, it will have smaller size. Zr(4d) ≈ Hf(5d), because of Lanthanide contraction.
Zn > Cu, their occur greater interelectronic repulsions in completely filled electronic configuration of 12 th group elements.
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