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4 years ago

Which of the following is a possible set of quantum numbers for an electron (n,l,m0,ms) ?

Chemistry

1 answer:

Mashutka [201]4 years ago

5 0

Answer is: (3, 2, 0, -1/2).

The principal quantum number (n) is one of four quantum numbers which are assigned to each electron in an atom to describe that electron's state.

For principal quantum number n=3:

1) azimuthal quantum number (l) can be l = 0...n-1:

l = 0, 1, 2.

The azimuthal quantum number determines its orbital angular momentum and describes the shape of the orbital.

2) magnetic quantum number (ml) can be ml = -l...+l.

ml = -2, -1, 0, +1, +2.

Magnetic quantum number specify orientation of electrons in magnetic field and number of electron states (orbitals) in subshells.

3) the spin quantum number (ms), is the spin of the electron.

ms = +1/2, -1/2.

(1, 1, 0, +1/2) is not correct because orbital quantum number cannot be l = 1 for n = 1.

(2, 1, 2, +1/2) is not correct because magnetic quantum number cannot be ml = 2 for orbital quantum number l = 1.

(3, -2, 1, -1/2) is not correct because orbital quantum number cannot be l = -2 for principal quantum number n = 3.

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Carbons starting from the left end:

sp²
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Refer to the sketch attached.

<h3>Explanation</h3>

The hybridization of a carbon atom depends on the number of electron domains that it has.

Each chemical bond counts as one single electron domain. This is the case for all chemical bonds: single, double, or triple. Each lone pair also counts as one electron domain. However, lone pairs are seldom seen on carbon atoms.

Each carbon atom has four valence electrons. It can form up to four chemical bonds. As a result, a carbon atom can have up to four electron domains. It has a minimum of two electron domains, with either two double bonds or one single bond and one triple bond.

A carbon atom with four electron domains is sp³ hybridized;
A carbon atom with three electron domains is sp² hybridized;
A carbon atom with two electron domains is sp hybridized.

Starting from the left end (H₂C=CH-) of the molecule:

The first carbon has three electron domains: two C-H single bonds and one C=C double bond; It is sp² hybridized.
The second carbon has three electron domains: one C-H single bond, one C-C single bond, and one C=C double bond; it is sp² hybridized.

The third carbon has three electron domains: two C-C single bonds and one C=O double bond; it is sp² hybridized.
The fourth carbon has two electron domains: one C-C single bond and one C≡C triple bond; it is sp hybridized.
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3 years ago

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xeze [42]

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2 years ago

One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K

mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ( $s_{2} - s_{1}$ ), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is:

$s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}$

$s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}$

The change in entropy is -1083.112 joules per kilogram-Kelvin.

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3 years ago