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m_a_m_a [10]
3 years ago
12

In the molecule ClF3, chlorine makes three covalent bonds. Therefore, three of its seven valence electrons need to be unpaired.

The orbitals with the same energy are known as degenerate orbitals. For example, the p subshell has three degenerate orbital, namely, px, py, and pz. How many degenerate orbitals are needed to contain seven electrons with three of them unpaired
Chemistry
1 answer:
Kisachek [45]3 years ago
3 0

Answer:

Explanation:

Chlorine has electronic configuration of 2 , 8 , 7

In n = 3 there are 7 electrons out of which 2 are in s , and 5 are in p . But out of 5 electrons in p , one electron jumps into d orbital . so the electronic configuration  becomes as follows

3s^23p_x^23p_y^23p_z^1  = 7

3s^23p_x^23p_y^13p_z^13d_{xy}^1

These orbitals like sp³d hybridise to form 7 degenerate orbitals out of which 2 orbitals contain electrons in pairs and rest three are singly occupied by electrons.( unpaired electrons )

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b) Calculate <span>Kb</span> for ephedrine.

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3 0
3 years ago
How many milliliters of nitrogen, N2, would have to be collected at 99.19 kPa and 28oC to have a sample containing 0.015 moles o
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Answer:

378mL

Explanation:

The following data were obtained from the question:

Pressure (P) = 99.19 kPa

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Next, we shall convert the pressure and temperature to appropriate units. This is illustrated below:

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For Temperature:

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T(°C) = 28°C

T(K) = 28°C + 273 = 301K.

Next we shall determine the volume of N2. The volume of N2 can be obtained by using the ideal gas equation as shown below:

PV = nRT

Pressure (P) = 0.98 atm

Temperature (T) = 301K

Number of mole (n) = 0.015 mole

Gas constant (R) = 0.0821atm.L/Kmol.

Volume (V) =...?

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Divide both side by 0.98

V = (0.015 x 0.0821 x 301) /0.98

V = 0.378 L

Finally, we shall convert 0.378 L to millilitres (mL). This is illustrated below:

1L = 1000mL

Therefore, 0.378L = 0.378 x 1000 = 378mL

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