The orbital having zero probability of finding electron on the surface of nucleus is and why 1. s 2. p_x 3. {d_{x^{2_{}}}}_{-y^2} 4. both 2 and 3

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Question 03

The probability density plots of 1s and 2s orbital are given in figure.

The density of dots in region represents the probability density of finding electrons in the region. On the basis of above diagram which of the following statements is incorrect ?

(a) 1s and 2s orbitals are spherical in shape

(b) The probability of finding electron is maximum near the nucleus

(d) The probability density of electrons for 2s orbital decreases uniformly as distance from the nucleus increases

Q. The probability of finding the electron in

p_{x} -

orbital is:

Q. The wave function for

1 s

orbital of hydrogen atom is given by:

ψ_{1 s} = \frac{π}{\sqrt{2}} e^{- r / a_{0}}

where,

a_{0} =

Radius of first Bohr orbit

r =

Distance from the nucleus (Probability of finding the electron varies with respect to it)
What will be the ratio of probabilities of finding the electrons at the nucleus to first Bohr's orbit

a_{0}

Q. Assertion :An electron in an

s

-orbital has a non-zero probability of being found right at the nucleus. Reason:

l = 0

for

s

-orbitals and thus there is no orbital angular momentum to fling the electron away from the nucleus.

Q. Select the incorrect option regarding

5 p_{x}

orbital.

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