Atomic or hybrid orbital on the central br atom makes up the sigma bond between this br and an outer f atom in bromine trifluoride, brf3 is sp2 hybridization
Trigonal hybridization is another name for sp2 hybridization. It entails combining one's' orbital with two 'p' orbitals of equal energy to create a new hybrid orbital known as sp2. A trigonal symmetry combination of s and p orbitals that is kept at 120
One of the hybrid orbitals formed when one s orbital and two p orbitals are mathematically merged to form three new equivalent orbitals orientated toward the corners of a triangle is sp2 hybridization.
The only feasible molecule geometry for sp2 hybridized center atoms is trigonal planar. When all of the bonds are in place, the shape is trigonal planar as well.
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Orbitals am only hold two electrons each, so 3 orbitals can hold 6 electrons
Answer: There are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series
Explanation:
Alpha Decay: In this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. The mass number is reduced by 4 units and atomic number is reduced by 2 units.
Beta Decay : It is a type of decay process, in which a proton gets converted to neutron and an electron. This is also known as -decay. In this the mass number remains same but the atomic number is increased by 1.
In radioactive decay the sum of atomic number or mass number of reactants must be equal to the sum of atomic number or mass number of products .
Thus for mass number : 235 = 207+4X
4X= 28
X = 7
Thus for atomic number : 92 = 82+2X-Y
2X- Y = 10
2(7) - Y= 10
14-10 = Y
Y= 4
Thus there are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series
Answer:
0.42%
Explanation:
<em>∵ pH = - log[H⁺].</em>
2.72 = - log[H⁺]
∴ [H⁺] = 1.905 x 10⁻³.
<em>∵ [H⁺] = √Ka.C</em>
∴ [H⁺]² = Ka.C
∴ ka = [H⁺]²/C = (1.905 x 10⁻³)²/(0.45) = 8.068 x 10⁻⁶.
<em>∵ Ka = α²C.</em>
Where, α is the degree of dissociation.
<em>∴ α = √(Ka/C) </em>= √(8.065 x 10⁻⁶/0.45) = <em>4.234 x 10⁻³.</em>
<em>∴ percentage ionization of the acid = α x 100</em> = (4.233 x 10⁻³)(100) = <em>0.4233% ≅ 0.42%.</em>
