Question

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Which of the following combinations of n and I represent real orbitals and which are impossible? Drag the appropriate items to their respective bins. Reset Help Q200 Real Impossible

Answer 1

Which of the following combination of n and l represent real arbital and which are impossible ? 45] 2015 25] Answer → Real- Impossible - 2 dl solution. Each orbital has some specific value of n and e. These value are called quantum Number. so for solving This problem, we first know the quantum Number. so there is type of quantum Number Principal quantum Number It is represented by n. It is directly propotional to size of orbital. It is the Integar value of arbital. i

•") Azimuthal quantum Number It is represent by e for calculating the value of l for any arbital, first you know the following Table Orbital value of l eslao 4 Remember All value of n and l is not bossible. There is some condition for real orbital. Condition for real orbital- & Value of n must be integar except zero and negative value. value of n must be greater than e. a w

In>l Now we solve the given problem. first we calculate value of n & l for all arbital. Than we check which orbital is not satisfying the ear condition of real orbital. 45] for 4s, value of his 4 an value of e=0 (fors, d=0). since n is integar and value of n (:4) is greater than value of l Go). Insel for 48 since real 4s follow all the orbital. so 4s condition for arbital is real 2d] for 2d, value of n = 2 and value of l = 2 (for d d = 2). since value of n (= 2 4 is not greater than value of l (=2). [ n l] → Not true for 2 d orbital

the So since 2d orbital do not follow the condition of real orbital 2d is Imaginary for is, value of n=1 and value of l = 0 (fors, l=0). Since [nal True for is since is of real orbital. follow all the orbital. So is condition is real T2P) for 2p, value of n=2 and value of l=1 (for P, l=1) since n>e] True for 2P since 2, follow all the condition of real orbital so 2p is real orbital
summary:

Real orbital: 4s , 1s , 2p

Imaginary orbital : 2d

Explanation:

For answering this problem, you first write the value of n and l for all orbital.

For real orbital, there is some conditions.

Conditions for real orbital:

1) value of n must be positive integer .

2) value of n must be greater than value of l.

value of n > value of l

Since 4s , 1s and 2p follow all the conditions of real orbital. So these orbitals are real.

For 2d , n= 2 and l=2

So for 2d , value of n is not greater than value of l. So 2d is imaginary orbital.