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NEETChemistryStructure of Atom

Match the following species with the correct number of electrons present in them: Species Number of Electrons   ( i )  Be 2 + ( a )   0   ( ii )  H + ( b )   10   ( iii )  Na + ( c )   2   ( iv )  Mg + ( d )   11 ( e )   4

Options

  1. A( i - d ) , ( i   i - c ) , ( i   i - b ) , ( i   v - a )
  2. B( i - a ) , ( i   i - b ) , ( i   i   i - c ) , ( i   v - d )
  3. C( i - e ) , ( i   i - d ) , ( i   i - a ) , ( i   v - c )
  4. D( i - c ) , ( i   i - a ) , ( i   i   i - b ) , ( i   v - d )

Correct answer

D. ( i - c ) , ( i   i - a ) , ( i   i   i - b ) , ( i   v - d )

Step-by-step solution

We know that the number of electrons is equal to the atomic number of element in neutral state. Atomic number of beryllium is 4 but Be 2 + can be formed by losing the two electrons so remaining electrons in Be 2 + will be 4 - 2   =   2 . Similarly, the number of electrons in H + is 1 - 1 = 0 . Similarly, the number of electrons in Na + is 11 - 1 = 10 . Similarly, the number of electrons in Mg + is 12 - 1 = 11 . Hence, the correct match will be ( i - c ) ,   ( ii - a ) ,   ( iii - b ) ,   (

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