(2014?临沂二模)已知A、B、C、D均为前四周期元素且原子序数依次增大,元素A的基态原子2p轨道有3个未成

2025-03-16 16:23:20
推荐回答(1个)
回答1:

A、B、C、D均为前四周期元素且原子序数依次增大,元素A的基态原子2p轨道有3个未成对电子,核外电子排布为1s22s22p3,则A为N元素;元素B的原子最外层电子数是其内层电子数的3倍,原子只能有2个电子层,最外层电子数为6,则B为O元素;元素C的一种常见单质为淡黄色粉末,则C为S元素;D的内层轨道全部排满电子,且最外层电子数为l,原子序数大于S,只能处于第四周期,核外电子数=2+8+18+1=29,则D为Cu元素,
(1)同周期随原子序数增大,元素第一电离能呈增大趋势,但氮元素原子2p轨道容纳3个电子,为半满稳定状态,能量较低,第一电离能高于O元素,高于第二周期中N、F、Ne的第一电离能都高于O元素,
故答案为:3;
(2)A的最简单气态氢化物为NH3,分子的空间构型为三角锥形;水分子与乙醇分子之间形成氢键,故H2O在乙醇中的溶解度大于H2S,
故答案为:三角锥形;水分子与乙醇分子之间形成氢键;
(3)NO3-中N原子价层电子对数=3+
5+1-2×3
2
=3,N原子轨道的杂化类型是sp2,与NO3-互为等电子体微粒的化学式为SO3、CO32-等,
故答案为:sp2;SO3、CO32-等;
(4)Cu(OH)2难溶于水,易溶于氨水,其溶于氨水的离子方程式为:Cu(OH)2+4NH3.H2O=[Cu(NH34]2++2OH-+4H2O,
故答案为:Cu(OH)2+4NH3.H2O=[Cu(NH34]2++2OH-+4H2O;
(5)Cu2O的晶胞中黑色球数目=4、白色球数目=2+8×
1
8
=2,故黑色球为Cu、白色球为O,晶胞质量=2×
64×2+16
NA
g,晶体的密度为ρ g?cm-3,则晶胞体积=
64×2+16
NA
g
ρ g?cm-3
=
288
ρ NA
cm3,则晶胞边长=
3
288
ρ NA
cm,
故答案为:
3
288
ρ NA

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