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ivann1987 [24]
3 years ago
14

Transition metal elements are in the "d block" on the periodic table, because the electron configuration of every transition met

al ends with electrons in "d" atomic orbitals. True False
Chemistry
2 answers:
goblinko [34]3 years ago
4 0
TRUE :) ;D;D;D;D;D ;J

MAVERICK [17]3 years ago
4 0

Answer:  TRUE

Explanation:  Transition elements are those which have electrons in the d orbitals in their outermost valence shell. They are also known as d block elements.

Thus Copper, Iron, Nickel etc are d block elements because their electrons are in the d orbital as shown-

Copper (Cu) = [Ar] 3d10 4s1

Iron (Fe) = [Ar] 3d6 4s2

Nickel (Ni) = [Ar] 3d8 4s2

Thus as shown in the electronic configuration above, the given stateent is true.

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The ionic radius for Na+ is 0.097 ηm and for Cl- is 0.181 ηm, the absolute value of the charge for each ion is 1.6x10-19 C, ε_o=
Vanyuwa [196]

Answer:

B = (2.953 × 10⁻⁹⁵) N.m⁹

Explanation:

At equilibrium, where the distance between the two ions (ro) is the sum of their ionic radii, the force between the two ions is zero.

That is,

Fa + Fr = 0

Fa = - Fr

Fa = (|q₁q₂|)/(4πε₀r²)

Fr = -B/(r^n) but n = 9

Fr = -B/r⁹

(|q₁q₂|)/(4πε₀r²) = (B/r⁹)

|q₁| = |q₂| = (1.6 × 10⁻¹⁹) C

(1/4πε₀) = k = (8.99 × 10⁹) Nm²/C²

r = 0.097 + 0.181 = 0.278 nm = (2.78 × 10⁻¹⁰) m

(k|q₁q₂|)/(r²) = (B/r⁹)

(k × |q₁q₂|) = (B/r⁷)

B = (k × |q₁q₂| × r⁷)

B = [8.99 × 10⁹ × 1.6 × 10⁻¹⁹ × 1.6 × 10⁻¹⁹ × (2.78 × 10⁻¹⁰)⁷]

B = (2.953 × 10⁻⁹⁵) N.m⁹

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