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The correct order of atomic/ionic radii is:
a) $Sc > Ti > V > Cr$
b) $Co > Ni > Cu > Zn$
c) ${S^{2 - }} > C{l^ - } > {O^{2 - }} > {N^{3 - }}$
d) None of these.

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Last updated date: 24th Jul 2024
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Answer
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Hint: We know that Atomic radius is by and large expressed similar to the absolute separation from a molecule's core to the furthest orbital of electrons. In less complex terms, it tends to be characterized as something like the range of a circle, where the focal point of the circle is the core and the external edge of the circle is the peripheral orbital of electrons. As you get across or down the intermittent table, patterns arise that help clarify how nuclear radii change.

Complete answer:
We must have to remember that the atomic radii of a synthetic component are a proportion of the size of its particles, normally the mean or regular separation from the focal point of the core to the limit of the encompassing shells of electrons. Since the limit is definitely not a clear cut actual element, there are different non-identical meanings of nuclear span. Four generally utilized meanings of nuclear span are: Van der Waals sweep, ionic range, metallic span and covalent sweep. Commonly, on account of the trouble to disconnect iotas to quantify their radii independently, nuclear range is estimated in a reinforced state; anyway hypothetical computations are obviously easier while thinking about molecules in separation. The conditions on climate, test, and state lead to a variety of definitions.
The correct order is,
$Sc > Ti > V > Cr$

Hence option A is correct.

Note:
We have to know that at the point when a covalent bond is available between two iotas, the covalent range can be resolved. At the point when two iotas of a similar component are covalently reinforced, the span of every particle will be a large portion of the distance between the two cores since they similarly draw in the electrons. The distance between two cores will give the measurement of a particle; however you need the span which is a large portion of the width.