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Ionic radii of (1) Ti^(4+) lt Mn^(7+) ...

Ionic radii of
(1) `Ti^(4+) lt Mn^(7+)`
(2) `""^(35)Cl^(-) lt ""^(37)Cl^(-)`
(3) `K^(+) gt Cl^(-)`
(4)`P^(5+) gt P^(3+)`
select the correct order

Text Solution

AI Generated Solution

To determine the correct order of ionic radii for the given ions, we will analyze each pair based on their charges and the number of protons and electrons they possess. ### Step-by-Step Solution: 1. **Compare `Ti^(4+)` and `Mn^(7+)`:** - Titanium (Ti) has an atomic number of 22, meaning it has 22 protons and, in the `Ti^(4+)` state, it has lost 4 electrons, resulting in 18 electrons. - Manganese (Mn) has an atomic number of 25, meaning it has 25 protons and, in the `Mn^(7+)` state, it has lost 7 electrons, resulting in 18 electrons. - Both ions have the same number of electrons (isoelectronic), but Mn has more protons (25) than Ti (22). The greater positive charge in Mn will pull the electrons closer, resulting in a smaller ionic radius. ...
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Ionic radii of (1) Ti^(4+) lt Mn^(7+) (2) ""^(35)Cl^(-) lt ""^(37)Cl^(-) K^(+) gt Cl^(-) Pt^(3+) gt P^(3+)

Ionic radii of (1) Ti^(4+) lt Mn^(7+) (2) ""^(35)Cl^(-) lt ""^(37)Cl^(-) K^(+) gt Cl^(-) Pt^(3+) gt P^(3+)

Knowledge Check

  • From amogst Al^(3+), Cl^(-), K^(+), N^(3-) , the correct order of radii is

    A
    `Al^(3+) lt N_(3-) lt K^(+) lt Cl_(-)`
    B
    `N^(3-) lt Al^(3+) lt Cl^(-) lt K^(+)`
    C
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    D
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  • Select the chain propagation steps in the free-radical chlorination of methane. (1) Cl_(2) to 2Cl^(*) (2) Cl^(*) + CH_(4) to CH_(3)Cl + H^(*) (3) Cl^(*) + CH_(4) to CH_(3)^(*) + HCl (4) H^(*) + Cl_(2) to HCl + Cl^(*) (5) CH_(3)^(*) + Cl_(2) to CH_(3)Cl + Cl^(*)

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    `2,3,5`
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