The hybridization for C in acetylene, HCCH, or C₂H₂ is 'sp'.
Discussion:
There are three different forms of hybridization -
- sp- The first occurs when two carbon atoms are triple linked.
- sp₂- When two carbon atoms are double-bonded to one another, this is known as sp₂.
- sp₃- When a single bond joins two carbon atoms, this is known as sp₃.
In the case of acetylene(HCCH or C₂H₂):
- The carbon atom requires additional electrons to establish four bonds with hydrogen and other carbon atoms in the synthesis of C₂H₂. One 2s₂ pair is consequently transferred to the vacant 2pz orbital. Each carbon has two sp hybrid orbitals after the 2s orbital in each atom combines with one of the 2p orbitals.
- As a result of the atoms' symmetrical alignment in a single plane, C₂H₂ possesses a linear molecular structure. Due to their lower electronegative nature than Hydrogen atoms, all Carbon atoms are situated near the center of the Lewis structure of C₂H₂.
H-C≡C-H
Therefore, it is concluded from the above discussion that the hybridization type of acetylene is 'sp'.
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Answer is: 12 grams of the isotope carbon-12 (¹²C).
The Avogadro’s number is the number of atoms in 12 grams of the isotope carbon-12 (¹²C).
Na is Avogadro number or Avogadro constant (the number of particles, in this example carbon, that are contained in the amount of substance given by one mole).
The Avogadro number has value 6.022·10²³ 1/mol in the International System of Units; Na = 6.022·10²³ 1/mol.
Answer:
1) the oxidation state of the metal is +2
2) there are four d electrons
3) metal valence electrons =4 ligand valence electrons= 18
4) sp^3d^2
5) octahedral geometry
Explanation:
[Cr(H2O)6]2+ is an octahedral complex. Octahedral complexes are known to have the metal ion in sp^3d^2 hybridization state. Since there are six ligands each bonding to the central metal ion, there are eighteen ligand valence electrons and four valence electrons from the metal present in the high spin complex.
The crystal field of high spin and low spin octahedral for a d4 ion is shown in the image attached.
The pressure of the hydrogen gas formed is given by its partial pressure at
the given temperature.
The pressure of the hydrogen gas formed is approximately <u>717.38 mmHg</u>.
Reasons:
The pressure of the hydrogen gas formed is given by the difference
between the vapor pressure of the gas collected and the vapor pressure of
water.
The vapor pressure of water at 24.0 °C is given by the Clausius-Clapeyron
Equation as follows;
Where;
P₁ = The partial pressure at temperature, T₁
P₂ = The pressure at temperature, T₂
We get;
The pressure of water at 24.0 °C, P₂ ≈ 22.62 mmHg
The vapor pressure of water at 24.0°C is 22.62 mmHg
Therefore;
The pressure of the hydrogen gas formed ≈ 740.0 mmHg - 22.62 mmHg = <u>717.38 mmHg</u>
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