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Gre4nikov [31]
4 years ago
15

According to valence bond theory, which orbitals on N and H overlap in the NH3 molecule? 2p on N overlaps with 2s on H 2s on N o

verlaps with 1s on H sp3 on N overlaps with 1s on H 2p on N overlaps with 1s on H sp3 on N overlaps with sp on H

Chemistry
1 answer:
Alecsey [184]4 years ago
8 0

Answer:

sp^{3} on N overlaps with 1s on H.

Explanation:

Hybridization of central atom generally can be found by finding the hybridization number.Hybridization number is the sum of number of lone pairs and number of sigma bonds.

If hybridization number is 2 then the central atom is sp hybridized.

If it is 3 it is sp^{2} hybridized.

If it is 4 it is sp^{3} hybridized and so on.

Here in NH_{3} molecule there are 3 sigma bonds and one lone pair(see the figure attached).

Hence hybridization number is 4 and hence it is sp^{3} hybridized.

And in case of hydrogen atom the only electron of it resides in the 1s orbital.

Therefore sp^{3} on N overlaps with 1s on H.

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3 years ago
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What would be the mass in grams of 8.94 x 10 22 formula units of copper(II) fluoride, CuF 2
Nadya [2.5K]

The mass in grams of 8.94×10²² formula units of CuF₂ is 15.07 g

<h3>Avogadro's hypothesis </h3>

6.02×10²³ formula units = 1 mole of CuF₂

But

1 mole of CuF₂ = 101.5 g

Thus,

6.02×10²³ formula units = 101.5 g of CuF₂

<h3>How to determine the mass of 8.94×10²² formula units of CuF₂</h3>

6.02×10²³ formula units = 101.5 g of CuF₂

Therefore,

8.94×10²² formula units = (8.94×10²² × 101.5 ) / 6.02×10²³

8.94×10²² formula units = 15.07 g of CuF₂

Thus, 8.94×10²² formula units are present in 15.07 g of CuF₂

Learn more about Avogadro's number:

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A student has a remote control for a television. The student points the
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In an experiment, a solution required 30. 05 g of nacl, 50. 0 g of , and 0. 4006 g of mgso4. Using the correct number of signifi
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When the use of significant figures and rounding up is applied correctly the mass of the mixture will be 80.5 g.

In cases of addition or subtraction, only the last significant figure of every number is taken into account.

In 30.05, this is 5, in the hundredths. When we look at 50.0, the last significant figure is 0, and it is in the tenths. And in 0.4006, the last significant figure is 6, in the ten thousandths. Of these three, the 0 from 50.0 is in the leftmost position, which means that the last significant figure of the result needs to be in the same position (in the tenths).

Moving onto the actual algebraic operation:

30.05 g + 50.0 g + 0.4006 = 80.4506 g

As we established, the last significant figure should be in the tenths, and we will have to round up 4 to 5 (trailing numbers are greater than 0), which means that the resulting mass will be 80.5 g.

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