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
- A( i - d ) , ( i   i - c ) , ( i   i - b ) , ( i   v - a )
- B( i - a ) , ( i   i - b ) , ( i   i   i - c ) , ( i   v - d )
- C( i - e ) , ( i   i - d ) , ( i   i - a ) , ( i   v - c )
- 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 ) ,   (