Show the distribution of electrons in oxygen atoms (atomic number 8) using an orbital diagram.

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Vishal Baghel | Contributor-Level 10

9 months ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans: The electronic configuration of oxygen atom is 1s² 2s2 2p⁴and its orbital diagram is given by

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Similar Questions for you

Find out the number of following orders which are INCORRECT against the mentioned properties:

(i) Ortho hydrogen > Para hydrogen (Stability at low temperature)

(ii) Ethanol > Glycerol (Viscosity)

(iii) D₂ < He (Boiling point)

(iv) HF > H₂O (Melting point)

(v) H₃BO₃ > BF₃ (Melting point)

(vi) NH₃ < SbH₃ (Boiling point)

(vii) H₂O₂ > H₂O (Strength of hydrogen bond)

(viii) H₂O < D₂O (Freezing point)

(ix) $H F_{2}^{Θ}$ < HF (Strength of hydrogen bond)

(x) D₂O < H₂O (Bond energy)

Vishal Baghel

Kindly go through the answers

(7.00)

Consider the following reaction:
(A substituted phenol reacts with H? and CrO?Cl to produce product 'P'.)
The product ‘P’ gives positive acetyl chloride test. This is because of the presence of which of these –OH group(s)?


Raj Pandey

Kindly consider the following Image

The number of radial land angular nodes in 4d orbital are, respectively

Payal Gupta

In 4d orbital, n = 4 and $l = 2$

Radial nodes = $n - l - 1$

Radial nodes = 4 – 2 – 1 = 1

And angular nodes, $l = 2$

The spin-only magnetic moment value of species is________× 10^-2 BM. (Nearest integer)

[Given:  $\sqrt[]{3} = 1.73]$

Alok Kumar Singh

$B_{2}^{+} \to σ 1 s^{2} s * 1 s^{2} σ 2 s^{2} σ * 2 s^{2} π 2 p^{1}$

Here, number of unpaired electrons, n = 1

Spin only moment ; $μ = \sqrt[]{n (n + 2)} B . M$

= 173 × 10^-2 B.M

Vishal Baghel

$Δ H = Δ U + Δ (P V)$

= $Δ U + P Δ V + V Δ P$

= $Δ U + P Δ V$ (At constant pressure)

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