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a. Compare the size of Cl, Cl^(ɵ) and Fe...

a. Compare the size of `Cl, Cl^(ɵ)` and `Fe^(2+)` ion.
b. the radii of `Ar` is greater than the radii of chlorine. Explain.

Text Solution

Verified by Experts

a. `(Z)/(e^(-))` ratio for `Cl = (17)/(17) = 1.00`
`Cl^(ɵ) = (17)/(18) = 0.944`
`Fe^(2+) = (26)/(24) = 1.08`
`Cl^(ɵ) gt Cl gt Fe^(2+)`
b. Explanation: In chlorine, the radii mean the atomic or covalent radii which is actually half the intermolecular distance between two atoms, whereas in argon the radii mean the van der Waals radii as argon is not a diatomic molecule. van der Waals radii is actually half the distance between adjacent molecule. So, van der Waals radii being larger than the atomic radii. Hence, argon has a larger radii than chlorine.
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Knowledge Check

  • If the bond distance in chlorine molecule (Cl_2) is 198 pmthen the radius of chlorine is

    A
    198pm
    B
    49.5pm
    C
    99pm
    D
    24.75pm

  • Consider the ioselectronic series , K^(o+), S^(2-), Cl^(ɵ), Ca^(2+) , the radii of the ions decrease as

    A
    `Ca^(2+) gt K^(o+) gt Cl^(ɵ) gt S^(2-)`
    B
    `Cl^(ɵ) gt S^(2-) gt K^(o+) gt Ca^(2+)`
    C
    `S^(2-) gt Cl^(ɵ) gt K^(o+) gt Ca^(2+)`
    D
    `K^(o+) gt Ca^(2+) gt S^(2-) gt Cl^(ɵ)`

  • Consider the isoelectronic series : K^(+), S^(2-) , Cl^(-) and Ca^(2+) , the radii of the ions decrease as

    A
    `Ca^(2+) gt K^(+) gt Cl^(-) gt S^(2-)`
    B
    `Cl^(-) gt S^(2-) gt K^(+) gt Ca^(2+)`
    C
    `S^(2-) gt Cl^(-) gt K^(+) gt Ca^(2+)`
    D
    `K^(+) gt Ca^(2+) gt S^(2-) gt Cl^(-)`

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    Assertion (A): Beryllium compounds are covalent in nature. Reason (R ): The size of Be^(2+) ion is larger in compariso to the radii of the other divalent ions of alkaline earth metals.

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