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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

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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+)

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

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The correct code for stability, of oxidation states for given cations is: (i) Pb^(2+) gt Pb^(4+) ,Tl^(+) lt Tl^(3+) (ii) Bi^(3+) lt Sb^(3+) , Sn^(3+) lt Sn^(4+) (iii) Pb^(3+) gt Pb^(4+),Bi^(3+) gt Bi^(3+) (iv) Tl^(3+) lt ln^(3+), Sn^(2+) gt Sn^(4+) (v) Sn^(2+) lt Pb^(2+) , Sn^(4+) gt Pb^(4+) (vi) Sn^(2+) lt Pb^(2+), Sn^(4+) lt Pb^(4+)

Consider the following orders: (i) HF > HCl > HBr > HI : Acidic strength (ii) CH_2 lt "CCl"_4 gt CF_4 : Electronegativity of central 'C' - atom (iii) Mg^(2+) lt K^(+) lt S^(2-) lt Se^(2-) : Ionic radius (iv) Ni > Pd > Pt : Ionic radius (v) As^(5+) gt Sb^(5+) gt BI^(5+) : Stable oxidation state (vi) Li^(+) lt Mg^(2+) lt Al^(3+) : Hydration energy (vii) Cl > Br > F > I : Electron affinity How many of these are correct orders?

Which of the following arrangements does not represent the correct order of the property stated against it ? (i) Ni^(2+) lt Co^(2+) lt Fe^(2+) lt Mn^(2+) : ionic size (ii) Co^(3+) lt Fe^(3+) lt Cr^(3+) lt Sc^(3+) : stability in aqueious solution (iii) Sc lt Ti lt Cr lt Mn : number of oxidation states (iv) V^(2+) lt Cr^(2+) lt Mn^(2+) lt Fe^(2+) : paramagnetic behaviour

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
    `Al^(3+) lt K^(+) lt Cl^(-) lt N^(3-)`
    D
    `K^(+) lt Cl^(-) lt Al^(3+) lt N^(3-)`
  • 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^(*)

    A
    `2,3,5`
    B
    `1,3,6`
    C
    `3,5`
    D
    `2,3,4`
  • Order H_(3)PO_(4) gt H_(2)SO_(4) gt HNO_(3) gt HCl is correct for :-

    A
    Acidic nature
    B
    Basic strength
    C
    Viscosity
    D
    Oxidation state
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