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Structure of Atom question

2005 · Shift 0 · Q32
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Structure of Atom question

2005 · Shift 0 · Q32

JEE MainChemistryStructure of AtomMCQ+4 / −1
Pick out the isoelectronic structure from the following : (i)      CH3+(ii)    H3O+(iii)   NH3(iv)    CH3−\begin{aligned} & \left( i \right)\,\,\,\,\,\,C{H_3}^ + \\ & \left( {ii} \right)\,\,\,\,{H_3}{O^ + } \\ & \left( {iii} \right)\,\,\,N{H_3} \\ & \left( {iv} \right)\,\,\,\,C{H_3}^ - \\\end{aligned}​(i)CH3​+(ii)H3​O+(iii)NH3​(iv)CH3​−​
  1. A
    (i)(i)(i) and (ii)(ii)(ii)
  2. B
    (iii)(iii)(iii) and (iv)(iv)(iv)
  3. C
    (i)(i)(i) and (iii)(iii)(iii)
  4. D
    (ii),(iii)(ii), (iii)(ii),(iii) and (iv)(iv)(iv)
View written solutionFree

Correct answer: D

  1. Meaning of isoelectronic

    Two or more species are isoelectronic if they have the same total number of electrons and usually a similar electronic arrangement.

  2. Count total electrons in each species

    We use atomic numbers:

    • C=6C = 6C=6
    • N=7N = 7N=7
    • O=8O = 8O=8
    • H=1H = 1H=1

  3. Species-wise calculation

    (i) CH3+CH_3^+CH3+​

    6+3(1)−1=86 + 3(1) - 1 = 86+3(1)−1=8 So, CH3+CH_3^+CH3+​ has 8 electrons.

    (ii) H3O+H_3O^+H3​O+

    8+3(1)−1=108 + 3(1) - 1 = 108+3(1)−1=10 So, H3O+H_3O^+H3​O+ has 10 electrons.

    (iii) NH3NH_3NH3​

    7+3(1)=107 + 3(1) = 107+3(1)=10 So, NH3NH_3NH3​ has 10 electrons.

    (iv) CH3−CH_3^-CH3−​

    6+3(1)+1=106 + 3(1) + 1 = 106+3(1)+1=10 So, CH3−CH_3^-CH3−​ has 10 electrons.

  4. Compare electron counts

    • CH3+→8CH_3^+ \rightarrow 8CH3+​→8 electrons
    • H3O+→10H_3O^+ \rightarrow 10H3​O+→10 electrons
    • NH3→10NH_3 \rightarrow 10NH3​→10 electrons
    • CH3−→10CH_3^- \rightarrow 10CH3−​→10 electrons

    Hence, the isoelectronic species are:

    H3O+, NH3, CH3−H_3O^+,\ NH_3,\ CH_3^-H3​O+, NH3​, CH3−​

  5. Match with options

    This corresponds to Option D: (ii),(iii)(ii), (iii)(ii),(iii) and (iv)(iv)(iv).

  6. Comparison with stored answer

    Stored correct answer = D

    Our derived answer = D

    So, they agree.

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