Period 4 transition element that forms 2+ ion with a half‐filled d sub level is
Manganese (Mn)
What is the half-filled d sub-level?
Transition metals are an interesting and challenging group of elements. They have perplexing patterns of electron distribution that don’t always follow the electron-filling rules. Predicting how they will form ions is also not always obvious.
Transition metals belong to the d block, meaning that the d sublevel of electrons is in the process of being filled with up to ten electrons. Many transition metals cannot lose enough electrons to attain a noble-gas electron configuration. In addition, the majority of transition metals are capable of adopting ions with different charges. Iron, which forms either the Fe2+ or Fe3+ ions, loses electrons as shown below.
Some transition metals that have relatively few d electrons may attain a noble-gas electron configuration. Scandium is an example. Others may attain configurations with a full d sublevel, such as zinc and copper.
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The formula to be used for this problem is as follows:
E = hc/λ, where h is the Planck's constant, c is the speed of light and λ is the wavelength. Also 1 aJ = 10⁻¹⁸ J
0.696×10⁻¹⁸ = (6.62607004×10⁻³⁴ m²·kg/s)(3×10⁸ m/s)/λ
Solving for λ,
λ = 2.656×10⁻⁷ m or <em>0.022656 nm</em>
Answer: Volume would be 196.15 mL if the temperature were changed to
and the pressure to 1.25 atmospheres.
Explanation:
Given:
,
= 256 mL,
= 720 torr (1 torr = 0.00131579 atm) = 0.947368 atm
,
Formula used to calculate volume is as follows.

Substitute the values into above formula as follows.

Thus, we can conclude that the volume would be 196.15 mL if the temperature were changed to
and the pressure to 1.25 atmospheres.
mole=10 x 10⁻³ : 46 g/mol = 2.17 x 10⁻⁴
<span>Answer:
Graham's law of gaseous effusion states that the rate of effusion goes by the inverse root of the gas' molar mass.
râšM = constant
Therefore for two gases the ratio rates is given by:
r1 / r2 = âš(M2 / M1)
For Cl2 and F2:
r(Cl2) / r(F2) = âš{(37.9968)/(70.906)}
= 0.732 (to 3.s.f.)</span>