<u>Answer:</u> The correct answer is Option D.
<u>Explanation:</u>
To calculate the hybridization of 
, we use the equation:
![\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20electron%20pair%7D%3D%5Cfrac%7B1%7D%7B2%7D%5BV%2BN-C%2BA%5D)
where,
V = number of valence electrons present in central atom (S) = 6
N = number of monovalent atoms bonded to central atom = 0
C = charge of cation = 0
A = charge of anion = 0
Putting values in above equation, we get:
![\text{Number of electron pair}=\frac{1}{2}[6]=3](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20electron%20pair%7D%3D%5Cfrac%7B1%7D%7B2%7D%5B6%5D%3D3)
The number of electron pair around the central metal atom are 3. This means that the hybridization will be 
and the electronic geometry of the molecule will be trigonal planar.
Hence, the correct answer is Option D.
Answer:
Fission. When a large fissile atomic nucleus such as uranium-235 or plutonium-239 absorbs a neutron, it may undergo nuclear fission. The heavy nucleus splits into two or more lighter nuclei, (the fission products), releasing kinetic energy, gamma radiation, and free neutrons.
Explanation:
The name of CuO is copper II oxide and its bond type is ionic or electrovalent bond.
<h3>What is electrovalent bond?</h3>
An ionic or electrovalent bond is the type of chemical bond where two atoms or molecules are connected to each other by electrostatic attraction.
This electrostatic attraction is as a result of the transfer of electrons from the metallic element to the non-metal.
According to this question, CuO is a chemical compound consisting of two elements namely; copper and oxygen. The compound name is copper II oxide.
Copper as a metal transfers electrons to oxygen atoms, hence, an ionic bond is formed between the molecules.
Learn more about ionic bonds at: brainly.com/question/11527546
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Answer:
There was an improvement in accuracy. There was no change in precision.
Explanation:
<em>The average mass after recalibration is closer to the mass of the standard, </em>so the recalibration improved the accuracy<em> </em>(the measurement is closer to an accepted 'true' value).
The standard deviation did not change, so the precision (or how disperse the measurements are) was not affected.
