JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer Consider the hydrates ions of \[T{{i}^{2+}},{{V}^{2+}},T{{i}^{3+}}\]and \[S{{c}^{3+}}.\]The correct order of their spin-only magnetic moments is : [JEE Main 10-4-2019 Morning]

    A) \[S{{c}^{3+}}<T{{i}^{3+}}<T{{i}^{2+}}<{{V}^{2+}}\]

    B) \[T{{i}^{3+}}<T{{i}^{2+}}<S{{c}^{3+}}<{{V}^{2+}}\]

    C) \[S{{c}^{3+}}<T{{i}^{3+}}<{{V}^{2+}}<T{{i}^{2+}}\]

    D) \[{{V}^{2+}}<T{{i}^{2+}}<T{{i}^{3+}}<S{{c}^{3+}}\]

    Correct Answer: A

    Solution :

    \[T{{i}^{+2}}=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{2}}\] unpaired electrons = 2. spin only magnetic moment\[(\mu )=\sqrt{2(2+2)}\]                                                 \[=\sqrt{8}B.M\] \[T{{i}^{+3}}=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{1}}\] unpaired electrons = 1 \[\mu =\sqrt{1(1+2)}=\sqrt{3}B.M\] \[{{V}^{+2}}=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{3}}\] \[\mu =\sqrt{3(3+2)}=\sqrt{15}B.M\] \[S{{c}^{+3}}=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}\] \[\mu =0\]


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