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人教版必修二《元素周期律》优秀教案

人教版必修二《元素周期律》优秀教案
人教版必修二《元素周期律》优秀教案

人教版必修二《元素周期律》优秀教案

必修2第一章物质结构元素周期律

第二节元素周期律(第一课时)

一、教材分析:

通过初三和必修I的学习,学生已经基本具备了一定的无机化学基础知识。例如初三学习的原子的构成、核外电子排布、元素周期表简介等一些基本的物质结构知识,这些为本章的学习奠定了一定的基础。在本章中,这些知识将更加细化,理论性更强,体系更加完整。通过《物质结构元素周期律》的学习,可以使学生对于所学元素化合物等知识进行综合、归纳。同时,作为理论指导,学生能更好的把无机化学知识系统化、网络化。在物质结构的基础上,将元素周期表的学习和元素周期律的学习结合起来,将学生在初中和必修I中所学习的氧化还原反应和许多元素化合物的知识连汇贯通。在第三节,通过化学键的学习,可以为今后有志深入学习化学的同学打下一定的基础。

总之,本章内容既是必修的重要理论内容,也是为选修内容的学习打下良好的基础。

二、教学目标:

知识目标:

1、知道元素原子核外电子排布规律;

2、掌握元素原子核外电子排布、原子半径、主要化合价周期性变化。

能力目标:

提高学生们归纳整理的能力。

情感、态度、价值观目标:

学会总结,学会概括,体会结构决定性质的理念。

三.教学重点难点:

重点:元素原子核外电子排布、原子半径、主要化合价周期性变化。

难点:知道元素原子核外电子排布规律。

四、学情分析:

通过初三和必修I的学习,学生已经基本具备了一定的无机化学基础知识。例如初三学习的原子的构成、核外电子排布、元素周期表简介等一些基本的物质结构知识,这些为本章的学习奠定了一定的基础。

五、教学方法:学案导学

六、课前准备:

学生学习准备:导学案

教师教学准备:投影设备

七、课时安排:一课时

八、教学过程:

(一)、检查学案填写,总结疑惑点(主要以学生课前板书答案展示的方式)

(二)、情景导入,展示目标:

老师:原子结构大家在初中就已经了解了,刚才通过同学们对导学案的展示已经对原子结构了解的不成问题了,那么核外电子具体怎样排

布?有什么规律特点,以及元素周期表能够体现出哪些结构规律,与元素性质规律有什么联系,是我们这节课要探讨的内容。

(三)、合作探究,精讲点拨

探究一:核外电子排布规律

教师:阅读课本P13表1-2,小结出原子核外电子排布规律思考以下问题:

展示投影:

(1)各电子层最多容纳_______个电子;

(2)最外层电子数不超过____个电子(K层为最外层时不超过_____个);

(3)次外层电子数不超过______个电子;

(4)核外电子总是尽先排布在能量____的电子层,然后由里向外,依次排布在能量逐步_____的电子层。

分组讨论:

1、写出1至20号元素的原子结构示意图.(找同学板书)

2、总结1至18号原子结构的特殊性。

(1)原子中无中子的原子:H11

(2)最外层有1个电子的元素H、Li、Na

(3)最外层有2个电子的元素:He、Be、Mg

(4)最外层等于次外层电子数的元素:Be、Ar

(5)最外层电子数是次外层电子数2倍的元素:C

(6)最外层电子数是次外层电子数3倍的元素:O

(7)最外层电子数是次外层电子数4倍的元素:Ne

※(8)氩原子电子层结构相同的阳离子是:K+;Ca 2+

与氩原子电子层结构相同的阴离子是:S2-;Cl-

核外有10个电子的粒子:

分子:CH4;NH3;H2O;HF;Ne

阳离子:NH4+;H3O+;Na+;Mg2+;Al 3+

阴离子:O2-;F-;OH-

探究二、元素的原子结构和主要化合价及原子半径的变化规律

教师:请阅读P14科学探究部分,比较1-18号元素化合价和半径的有关数据,从中能找出什么规律?

学生:随着原子序数的递增,元素原子的电子层排布、原子半径和化合价都呈现周期性变化。

(Ⅰ)半径

讨论:(1)第二周期元素从左到右半径变化规律是什么?第三周期呢?为什么?

(2)比较O2-与O半径大小?Ca 2+与Ca原子半径大小?

(3)比较O2-与Na+半径大小?

学生:分小组讨论后给出各组答案.

最后总结:简单微粒半径的比较方法:

(1)电子层书相同时,随原子序数的递增,原子半径逐渐减小。

(2)最外层电子数相同时,随电子层数的递增原子半径逐渐增大。

(3)同种元素:阴离子大于原子,原子大于阳离子。

(4)电子层结构相同的离子,核电荷数越大半径越小。

(5)带相同电荷的离子,电子层数越多,半径越大。

总体方法:比较微粒半径大小,先看电子层数,电子层数越多半径越大;电子层数相同,再看核电荷数,核电荷数越多半径越小。

(Ⅱ)化合价

教师:价电子——元素原子的最外层电子或某些元素的原子的次外层或倒数第三层的部分电子。

投影:

1、主族元素的最高正化合价=______________

=______________

2、非金属最高正价+|负化合价|=______

(注:副族和第VⅢ族化合价较复杂)

学生阅读课本回答:最外层电子数主族的族序数8

教师:思考:金属的化合价有什么特点?记住氟没有正化合价.

学生:金属元素全是正价。

投影:填写表格;

(四)反思总结,当堂检测:

1、原子序数从11依次增加到17,下列递变关系中,错误的是()

A.电子层数逐渐增多B..原子半径逐渐增大

C.最高正化合价数值逐渐增大D.从Si到Cl,最低负化合价从-4到-1

2、下列各组元素中按微粒半径递增顺序排列的是()

A、Li Na K

B、Ba2+Ca2+Mg2+

C、Ca2+K+Cl-

D、N O F

3、下列元素原子半径最大的是()

A、Li

B、F

C、Na

D、Cl

4、按原子半径增大的顺序排列的一组是()

A、Be、N、F

B、Mg、Si、C

C、Na、Mg、Ca

D、Cl、S、P

5、元素X原子的最外层有3个电子,元素Y原子的最外层有6个电子,这两种元素形成的化合物的化学式可能是()

A.XY2

B.X2Y3

C.X3Y2

D.X2Y

6、A、B两种原子,A原子M层比B原子M层少3个电子,B原子L 层电子数恰好是A原子L层的两倍,则A元素是()

A.O

B.Si

C.C

D.Mg

7、元素性质呈周期性变化的决定因素是()

A.元素原子半径大小呈周期性变化

B.元素原子量依次递增

C.元素原子最外层电子排布呈周期性变化

D.元素的最高正化合价呈周期性变化

8、a X n-和b Y m+两种简单离子,其电子层结构相同,下列关系式或化学式正确()

A、a–n=b+m

B、a+m=b–n

C、氧化物为YO m

D、氢化物为H n X或XH n

9、已知元素X能形成H2X和XO2化合物,则X的最外层电子数为

某元素气态氢化物的分子式为H2R,该元素的最高价氧化物的化学式为__.

10、原子序数为34的元素位于周期,族,属于类单质;原子序数为56的元素位于周期,族,属于类单质。

11、A、B、C、D、4种元素,已知A—核外有18个电子;B原子最外层电子数比D原子核外电子数多2个,B原子比D原子多2个电子层;D+核外没有电子,C元素原子核外电子数比B元素原子核外电子数多5个。(1)、写出四种元素的名称。(2)、画出C和D原子及A-的结构示意图。

(1)A、B、C、D

(2)

(五)发导学案,布置预习:略。

九、板书设计:

一、核外电子排布规律

二、元素的原子结构和主要化合价及原子半径的变化规律

三、简单微粒半径的比较方法

十、教学反思:

这节课采用提出问题——分组讨论——总结规律——归纳方法——练习巩固的方式让所有

同学都能积极参与进来。课堂气氛活跃,学习氛围浓厚,让同学们在真正的探索中学习。

十一、参考答案见学案

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